Projectile Motion Using Vectors

In summary: to shorten the range to the blue curve while maintaining the point ##(x_1,y_1)## you would have to increase both the velocity and the angle?
  • #1
themethetion
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Thread moved from the technical math forums to the schoolwork forums
TL;DR Summary: Using vector functions how can I find the minimum average velocity (something greater than 11.86m/s) of a ball if the launch angle is unknown and if I have a point that the object must travel through (11.86, 3.47)?

In my assignment, I developed a function for a lob shot (opposition against the net, the player hitting the ball from the back of the court) in tennis where the ball travels over a player's maximum reach (3.47m) and restriction in court length (23.72m). Therefore through points (11.86, 3.47). The function developed was 3.47= (-4.9/(cos^2(x))) + 11.86tan(x). This gave x/theta as imaginary numbers therefore I'm assuming with an average velocity of 11.86m/s (calculated from professionals performing a lob) the height and distance of 11.86mX and 3.47mY cannot be reached. Therefore I want to determine the minimum velocity and its corresponding launch angle that I can use for the tennis ball to travel through this point while landing before 23.72m. But I'm unsure of how to calculate the minimum velocity without a launch angle.
 
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  • #2
Why must the speed > 11.86 m/s??
 
  • #3
The general equation is:

$$y(x) = y_o + \tan \theta x - \frac{gx^2}{2 \cos^2 \theta v^2}$$

You then solve for ##v## in terms of the point that the projectile must pass through ##(x_1, y_1)##

You will then have a function ##v(\theta)##.

Next, attempt to find an optimum solution for ##v##, by taking the derivative of ##v## w.r.t ##\theta##. i.e. ##v'(\theta) = 0##

Make sure ##v## is optimized in the way you expect (minimized)by normal tests.
 
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  • #4
themethetion said:
TL;DR Summary: Using vector functions how can I find the minimum average velocity (something greater than 11.86m/s) of a ball if the launch angle is unknown and if I have a point that the object must travel through (11.86, 3.47)?

In my assignment, I developed a function for a lob shot (opposition against the net, the player hitting the ball from the back of the court) in tennis where the ball travels over a player's maximum reach (3.47m) and restriction in court length (23.72m).
This doesn't seem like a realistic solution to the problem. A player at the net is going to move backwards in response to a lob, rather than remain at the net.

Also, the ball is not hit from ground level, but from nearer to 1m off the ground. This is the ##y_0## in the post above.

That said, the problem of hitting the ball (from off the ground) to clear an obstacle and land within a specified range is an interesting one.
 
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  • #5
PeroK said:
That said, the problem of hitting the ball (from off the ground) to clear an obstacle and land within a specified range is an interesting one.
I missed this bit of info... I guess they should instead try to minimize the range given point ##(x_1,y_1)##?

EDIT: Actually never mind, I think trying to minimize the range would not work with the velocity remaining variable. If you found a particular solution, you could always increase the angle and velocity simultaneously to shorten the range.
 
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  • #6
I'm now thinking that any trajectory that satisfies ##(x_1,y_1)## and ## (R, 0)## (the range of the projectile is ##R##) then that must be the minimum velocity?

$$ v^2 = \frac{ gR^2 - \sin ( 2 \theta ) R}{2 y_o \cos^2 \theta}$$

and

$$ y_1 = y_o + \tan \theta x_1 - \frac{g {x_1}^2}{ \cos^2 \theta v^2}$$

So if you substitute for ##v^2## , ##\theta## is fixed by the solution of the resulting equation.

If you desired to shorten the range while maintaining ##(x_1,y_1)##, you must increase both ##v## and ##\theta## to do so. It seems like if you find a solution to what is stated above it seems like it must be the optimal solution which minimizes the velocity?
 
  • #7
Three points determine a parabola. You need some flexibility in the points to have a minimisation problem.
 
  • #8
PeroK said:
Three points determine a parabola. You need some flexibility in the points to have a minimisation problem.
To me it seems like the solution would necessarily have to be a minimum though?

1676474262804.png


If ##R## is the maximum possible range the trajectory can have while passing through point ##(x_1,y_1)## ( the red trajectory), then if we desired to shorten the range to the blue curve while maintaining ##( x_1,y_1)##, we would have to increase the velocity ( and angle ) to do so (the blue trajectory)?

So what I'm saying is scratch the minimization problem in post #3, its not going to work.
 
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  • #9
erobz said:
To me it seems like the solution would necessarily have to be a minimum though?

If ##R## is the maximum possible range the trajectory can have while passing through point ##(x_1,y_1)## ( the red trajectory), then if we desired to shorten the range to the blue curve while maintaining ##( x_1,y_1)##, we would have to increase the velocity ( and angle ) to do so (the blue trajectory)?

So what I'm saying is scratch the minimization problem in post #3, its not going to work.
This can't be true in general. Although, it's not clear precisely what assumptions you are making.
 
  • #10
PeroK said:
This can't be true in general. Although, it's not clear precisely what assumptions you are making.
Can you show me a counter example, because I can't see what I'm missing?
 
  • #11
It looks like your problem can be recast into the following formulation: Find the minimum initial speed required for a projectile to travel horizontal and vertical displacements ##\Delta x## and ##\Delta y## respectively. The trajectory equation in post #3 can be rewritten as $$\Delta y= \tan \theta \Delta x - \frac{g(\Delta x)^2}{2 v_0^2 \cos^2 \theta} .$$ There are four variables, ##\Delta x##, ##\Delta y##, ##\theta## and ##v_0##. You know the first two, so you need an additional optimization equation. The equation to use is
$$\sqrt{(\Delta x)^2+(\Delta y)^2 }+\Delta y=\frac{v_0^2}{g}.$$ See this insight article for a derivation of this equation and for additional information, e.g. for the projection angle when ##v_0## is optimized. Also see Example 4 in that article which is similar to your problem.

Do you really mean "average velocity" or would "initial velocity" be more appropriate to what you are looking for?
 
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  • #12
Assumptions: There is a single parabola ##P## that satisfies ##(0,y_o), ( x_1,y_1), ( R, 0)##. The initial velocity ##v## and angle ##\theta## that satisfy ##P## are fixed. If we allow variation that ##R' \leq R##, then the velocity ##v'## and angle ##\theta'## must both be increased relative to ##v## and ##\theta## to maintain ##( 0,y_o), ( x_1,y_1)## with new third condition ##(R',0)##. Where does that go wrong?
 
  • #13
kuruman said:
It looks like your problem can be recast into the following formulation: Find the minimum initial speed required for a projectile to travel horizontal and vertical displacements ##\Delta x## and ##\Delta y## respectively. The trajectory equation in post #3 can be rewritten as $$\Delta y= \tan \theta \Delta x - \frac{g(\Delta x)^2}{2 v_0^2 \cos^2 \theta} .$$ There are four variables, ##\Delta x##, ##\Delta y##, ##\theta## and ##v_0##. You know the first two, so you need an additional optimization equation. The equation to use is
$$\sqrt{(\Delta x)^2+(\Delta y)^2 }+\Delta y=\frac{v_0^2}{g}.$$ See this insight article for a derivation of this equation and for additional information, e.g. for the projection angle when ##v_0## is optimized. Also see Example 4 in that article which is similar to your problem.

Do you really mean "average velocity" or would "initial velocity" be more appropriate to what you are looking for?
I don't think there is any optimization. I messed up in post 3. The parabola is fully determined by the given points:

The initial height at ##x = 0, ( 0,y_o)##
The obstacle position ##( x_1,y_1)##
The maximal range ##( R, 0 )## for it to remain in play while (just) clearing the obstacle.

If there is a solution, I feel like it is coming from post #6 from which ##\theta## and ##v## can be fully determined by:

$$y_1 = y_o + \tan (\theta ) x_1 - \frac{g x_1^2 y_o}{gR^2 - \sin ( 2 \theta) R}$$

Am I messing up?
 
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  • #14
erobz said:
I don't think there is any optimization. I messed up in post 3. The parabola is fully determined by the given points:
You are assuming that the minimum velocity corresponds to the furthest landing point. It doesn’t have to land on the baseline, just not go past it.
@themethetion, what assumption are you making about the height at which the ball is struck?
 
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  • #15
haruspex said:
You are assuming that the minimum velocity corresponds to the furthest landing point. It doesn’t have to land on the baseline, just not go past it.
@themethetion, what assumption are you making about the height at which the ball is struck?
Yeah, but if the total range is reduced, and the parabola must still pass through ##(x_1,y_1)##, then the initial velocity and angle must be increased as far as I can tell. This is what I was trying to get at with post #8.
 
  • #16
erobz said:
If ##R## is the maximum possible range the trajectory can have while passing through point ##(x_1,y_1)## ( the red trajectory),
There is no such maximum. Send it at light speed through that point.
You might instead have defined the red trajectory as the path of minimum launch speed that goes through the target point, but you cannot assume its landing point is beyond the baseline.
 
  • #17
haruspex said:
There is no such maximum. Send it at light speed through that point.
You might instead have defined the red trajectory as the path of minimum launch speed that goes through the target point, but you cannot assume its landing point is beyond the baseline.
No, ##R## is the maximum range for which the ball passes the obstacle ##(x_1,y_1)## and lands in bounds. They are talking about a tennis lob.
 
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  • #18
erobz said:
No, R is the maximum range for which the ball passes the obstacle and land in bounds.
That's not what you wrote in post #8.
Forget the baseline for the moment. There is a minimum speed at which the ball can be made to pass through the target point. That determines a landing point. You are assuming the baseline is nearer than that, but it could be further away. Either way, you would have to increase the launch speed to make it land on the baseline.
 
  • #19
haruspex said:
That's not what you wrote in post #8.
Forget the baseline for the moment. There is a minimum speed at which the ball can be made to pass through the target point. That determines a landing point. You are assuming the baseline is nearer than that, but it could be further away. Either way, you would have to increase the launch speed to make it land on the baseline.
Ahh see the error in my assumption. It could be that it makes point ##(x_1,y_1)## and lands in bounds. Thank You.

Its still not clear what is being optimized here. If the minimum velocity land outside of bounds, we have to increase the velocity and angle until it lands in bounds. However, if it lands in bounds at minimum velocity then we take that solution. Is that correct?
 
  • #20
erobz said:
Ahh see the error in my assumption. It could be that it makes point ##(x_1,y_1)## and lands in bounds. Thank You.

Its still not clear what is being optimized here. If the minimum velocity land outside of bounds, we have to increase the velocity and angle until it lands in bounds. However, if it lands in bounds at minimum velocity then we take that solution. Is that correct?
Yes.
 
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  • #21
erobz said:
I don't think there is any optimization. I messed up in post 3. The parabola is fully determined by the given points:

The initial height at ##x = 0, ( 0,y_o)##
The obstacle position ##( x_1,y_1)##
The maximal range ##( R, 0 )## for it to remain in play while (just) clearing the obstacle.

If there is a solution, I feel like it is coming from post #6 from which ##\theta## and ##v## can be fully determined by:

$$y_1 = y_o + \tan (\theta ) x_1 - \frac{g x_1^2 y_o}{gR^2 - \sin ( 2 \theta) R}$$

Am I messing up?
Since optimization is not the issue and you want to know the possible ranges with the projectile just clearing point ##(x_1,y_1)##, then equation (4) in the article I quoted in post #11 is the way to go. It is the trajectory equation transformed to include the tangent of the projection angle ##\theta## and the tangent of the angle ##\varphi## that the instantaneous position vector makes relative to the horizontal. It is extremely useful, especially with problems like this one. $$\frac{\Delta x}{\tan\!\theta-\tan\!\varphi}=\frac{2v_{0x}^2}{g}.\tag{1}$$ Note that the right-hand side is a constant of the motion which means that the left-hand side is also a constant of the motion. This eliminates the need for ##v_0## because we can write $$\frac{\Delta x_1}{\tan\!\theta-\tan\!\varphi_1}=\frac{\Delta x_2}{\tan\!\theta-\tan\!\varphi_2}$$Using names of variables as you defined them, we have
##\Delta x_1=x_0##, ##~\Delta x_2=R##; ##~\tan\!\varphi_1=\dfrac{y_1-y_0}{x_0}##; ##~\tan\!\varphi_2=\dfrac{-y_0}{R}##. If one puts them in the equation above, one gets a quadratic in ##R## with ##\tan\!\theta## the independent variable. It can be solved to yield the range as a function of projection angle given the constraint that the projectile passes through point ##(x_1,y_1)##. The solution produces pairs of projection angles and ranges. The initial speed is strictly determined, given a choice of the projection angle, using equation (1).

I generated two plots using parameters other than OP's (for obvious reasons). I chose ##x_0=16~## m; ##~y_0=2~##m;##~y_1=4~##m and generated a first plot giving the range as a function of projection angle. At each projection angle there is only one value of the initial speed such that the object passes through ##(x_1,y_1)=(16,4)~##m. If ##R## is known, then there is only one value for the projection angle and, through equation (1) above, only one value for ##v_0.## This first plot is shown below.
Range.png

Then I plotted two trajectories by choosing angles and corresponding speeds. I chose
Plot A (red): ##\theta = 11^o##; ##~v_0=33.2~##m/s
Plot B (blue): ##\theta = 32^o##; ##~v_0=12.7~##m/s
The trajectories intersect at ##(16,4)~##m as expected and are shown in the second plot below.
Trajectory.png
 
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  • #22
hutchphd said:
Why must the speed > 11.86 m/s??
I calculated a balls average velocity assuming it’s linear using a video. This failed and i ended up solving for it simultaneously with theta.
 
  • #23
PeroK said:
This doesn't seem like a realistic solution to the problem. A player at the net is going to move backwards in response to a lob, rather than remain at the net.

Also, the ball is not hit from ground level, but from nearer to 1m off the ground. This is the ##y_0## in the post above.

That said, the problem of hitting the ball (from off the ground) to clear an obstacle and land within a specified range is an interesting one.
I was under the impression because i’m assuming the ball isnt hit from the ground and it is actually hit from approximately waist level. but the calculations assume it is hit from the ground, using the same angle and magnitude would result in the ball would clearing the opposing player by the amount not yet considered.
 
  • #24
haruspex said:
You are assuming that the minimum velocity corresponds to the furthest landing point. It doesn’t have to land on the baseline, just not go past it.
@themethetion, what assumption are you making about the height at which the ball is struck?
It must land close to the baseline, as it must be as difficult as possible for the opposing player to return the shot. This is also why it is being hit onto the players off-hand(furthest right hand side of court).
The calculations assume the ball is hit from the ground but i’m assuming the ball is hit from waist level. Therefore i thought the ball would travel from waist level to the ground above the opposing player if not taken into consideration. is that right?
 
  • #25
themethetion said:
It must land close to the baseline
Then you need decide what exactly the criteria are.
How about: it must land on the baseline; it must be above what the opponent can reach; subject to those, find the minimum velocity?
 
  • #26
kuruman said:
Since optimization is not the issue and you want to know the possible ranges with the projectile just clearing point ##(x_1,y_1)##, then equation (4) in the article I quoted in post #11 is the way to go. It is the trajectory equation transformed to include the tangent of the projection angle ##\theta## and the tangent of the angle ##\varphi## that the instantaneous position vector makes relative to the horizontal. It is extremely useful, especially with problems like this one. $$\frac{\Delta x}{\tan\!\theta-\tan\!\varphi}=\frac{2v_{0x}^2}{g}.\tag{1}$$ Note that the right-hand side is a constant of the motion which means that the left-hand side is also a constant of the motion. This eliminates the need for ##v_0## because we can write $$\frac{\Delta x_1}{\tan\!\theta-\tan\!\varphi_1}=\frac{\Delta x_2}{\tan\!\theta-\tan\!\varphi_2}$$Using names of variables as you defined them, we have
##\Delta x_1=x_0##, ##~\Delta x_2=R##; ##~\tan\!\varphi_1=\dfrac{y_1-y_0}{x_0}##; ##~\tan\!\varphi_2=\dfrac{-y_0}{R}##. If one puts them in the equation above, one gets a quadratic in ##R## with ##\tan\!\theta## the independent variable. It can be solved to yield the range as a function of projection angle given the constraint that the projectile passes through point ##(x_1,y_1)##. The solution produces pairs of projection angles and ranges. The initial speed is strictly determined, given a choice of the projection angle, using equation (1).

I generated two plots using parameters other than OP's (for obvious reasons). I chose ##x_0=16~## m; ##~y_0=2~##m;##~y_1=4~##m and generated a first plot giving the range as a function of projection angle. At each projection angle there is only one value of the initial speed such that the object passes through ##(x_1,y_1)=(16,4)~##m. If ##R## is known, then there is only one value for the projection angle and, through equation (1) above, only one value for ##v_0.## This first plot is shown below.
View attachment 322334
Then I plotted two trajectories by choosing angles and corresponding speeds. I chose
Plot A (red): ##\theta = 11^o##; ##~v_0=33.2~##m/s
Plot B (blue): ##\theta = 32^o##; ##~v_0=12.7~##m/s
The trajectories intersect at ##(16,4)~##m as expected and are shown in the second plot below.
View attachment 322337
I’m sorry i should have cleared up the fact we can’t use physics formulas as it is a math assignment. we can only integrate from acceleration to displacement. This is the procedure i ended up following but i believe i ran into errors as i must have some misconceptions.

I modified the acceleration of the ball so i could solve for the velocity and theta at the same time.

I planned to combine the magnitude and angle for x & y with the magnitude and angle for x & z into one vector equation. I now believe this won't work and am stuck on how to even create a 3d projectile motion vector while making it somewhat practical in the real life after developing a solution to help a tennis player.

I calculated the angle of projection and velocity of x & y by integrating them from acceleration with ax = 0m/s and ay= -9.8m/s to displacement
With the function, I ended up with I created the following vector equation. r_t = (xcos(θ)t)i + (-4.9t^2+xsin(θ)t)j where x is the velocity and theta is the angle of projection
I then let R_x= x cos(θ)t and R_y= -4.9t^2 + xsin(θ)t
R_x was then let equal to the maximum x value the ball can travel through
This was rearranged for t and subbed into R_y
R_y was then let equal to the corresponding y value.
The same process was followed to develop another equation. These equations were plotted into Desmos and the intersect was determined as (16.35, 0.515) therefore it was believed the ball must have a velocity of 16.35m/s and a projection of 0.515 radians to go through said points in the xy plane.
The same process was followed for x & z. where acceleration was integrated to find displacement from ax = 0m/s and az=9m/s (the average velocity of wind for complexity). But the end functions gave no real results where points, were used in the path the ball, must travel. Which made me question my original method.
When calculating for x & z should the velocity and/or the angle of projection of R_x= x cos(θ)t be that found when solving for x & y. hence should the equation be
R_x= 16.35 cos(θ)t
t= r_x/(16.35cos(θ))
R_y= -4.9(r_x/(16.35cos(θ))^2 + xsin(θ)(r_x/(16.35cos(θ))
or
t= r_x/(xcos(θ))
R_y = -4.9(r_x/(xcos(θ))^2 + xsin(θ)(r_x/(xcos(θ))

Please consider:
Values have been modified.
The ball must travel above 4mX at 4.5mZ ( the opposition), and the ball must land on the baseline 25mY on the furthest right-hand side of the court at 9mZ. These points are all with respect to the player, at the origin.
 
  • #27
haruspex said:
Then you need decide what exactly the criteria are.
How about: it must land on the baseline; it must be above what the opponent can reach; subject to those, find the minimum velocity?
Yes, but the velocity and launch angle of the ball for it to meet these points. See post #24 for the method i tried which failed.
The ball must travel above 4mX at 4.5mZ ( the opposition), and the ball must land on the baseline 25mY on the furthest right-hand side of the court at 9mZ. These points are all with respect to the player, at the origin.
 
  • #28
themethetion said:
With the function, I ended up with I created the following vector equation. r_t = (xcos(θ)t)i + (-4.9t^2+xsin(θ)t)j where x is the velocity and theta is the angle of projection
I then let R_x= x cos(θ)t and R_y= -4.9t^2 + xsin(θ)t
R_x was then let equal to the maximum x value the ball can travel through
This was rearranged for t and subbed into R_y
R_y was then let equal to the corresponding y value.
The same process was followed to develop another equation. These equations were plotted into Desmos and the intersect was determined as (16.35, 0.515) therefore it was believed the ball must have a velocity of 16.35m/s and a projection of 0.515 radians to go through said points in the xy plane.
The same process was followed for x & z.
So you found a launch angle and speed as though the court had zero width, then tried to adjust it by adding a lateral component to the velocity to allow for landing at the far side of the court? That's ok, but the launch angle will now be less.
Why not just take the x axis as running diagonally across the court?
 
  • #29
haruspex said:
So you found a launch angle and speed as though the court had zero width, then tried to adjust it by adding a lateral component to the velocity to allow for landing at the far side of the court? That's ok, but the launch angle will now be less.
Why not just take the x axis as running diagonally across the court?
I thought of this today, if you mean by looking at it as though it's 2D where the x-axis is diagonal across the court and then the y-component is the height.
How would this affect my calculations? I would need to change the value of x to the diagonal distance rather than the vertical distance and redo it, correct?
 
  • #30
themethetion said:
How would this affect my calculations?
It avoids the complication of the launch angle needing to change when you introduce the lateral motion,
themethetion said:
I would need to change the value of x to the diagonal distance rather than the vertical distance and redo it, correct?
Yes, and of the opponent’s position.
 
  • #31
haruspex said:
It avoids the complication of the launch angle needing to change when you introduce the lateral motion,

Yes, and of the opponent’s position.
But the opponent is on the diagonal, so wouldn't his position remain the same?
 
  • #32
themethetion said:
But the opponent is on the diagonal, so wouldn't his position remain the same?
The position on the court is the same, but not the distance from the origin.
 
  • #33
haruspex said:
The position on the court is the same, but not the distance from the origin.
But the origin hasn't changed?
 
  • #34
themethetion said:
But the origin hasn't changed?
In your method in post #26, you started off by ignoring the width of the court. You treated the trajectory as being straight up one side of the court, baseline to baseline, with the opponent half way.
Changing it to a corner to corner trajectory increases the distance from the server to the opponent as well as from server to landing point.
 
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  • #35
I am sure that you have the skills to work this out if you think clearly and work methodically, but after 34 posts you still seem to be struggling. I think there are two main reasons for the struggle. First you don't seem to have a mental image of the problem. The best way to get a mental image is to draw a diagram: do this and post it here, with everything labelled. Which brings me to the second problem: you have used non-standard and very confusing labels for things, in particular
themethetion said:
x is the velocity
Don't do this. Choose ## v ## for the initial velocity (actually it is speed, but let's stick with ## v ## because we all know we are talking about the magnitude of the velocity here), then you can use ## x ## instead of R_x and ## y ## instead of R_y. Now using ## g ## for gravity (always keep symbols in your workings, only substituting numbers in at the end to get an answer) we can rewrite
themethetion said:
R_x= x cos(θ)t and R_y= -4.9t^2 + xsin(θ)t
as ## x = (v \cos \theta) t ## and ## y = y_0 + (v \sin \theta) t - (\frac 1 2 g) t^2 ## (note I have also introduced ## y_0 ##, the initial height of the ball as I am sure the player is not hitting it off the ground). We can then use sensible labels for things like the distance to the baseline ## x_b ## and to the distance and height to clear the other player ## x_p ## and ## y_p ##. Mark these on your diagram.
 
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