Find the Initial Velocity and Launch Angle for a Particular Trajectory

In summary, the problem is to find the initial velocity and angle needed to launch a projectile and reach the coordinates (x,y), while also reaching a maximum height of y+5m before heading down to the target location. The given variables are the initial and final positions, mass of the ball, gravity, and maximum height. The time to reach the target is not necessary. The equations Vx = Vi*Cos(Angle) and Vy = Vi*Sin(Angle) will be used. The initial vertical velocity component can be used to calculate the time required to reach the target height, and then the time to reach the target can be found by adding the time to reach the maximum height. The final step would be to use the distance
  • #1
Francis Perez
7
0
Thread title changed to be more descriptive of the problem.

Homework Statement


Hello, so I’ve been trying to find some possible way to calculate this:
The initial Velocity and angle needed to launch a projectile and reach the coordinates (x,y). The maximum height the ball can reach before it starts to head down to the target value should be y+5m
The variables are only given by the initial place, being (0,0), the final place, be (X,Y), the mass of the ball, the gravity, maximum height, (Y+5). The time it will take there, I don’t know, it’s not necessary to find the time, just the angle and initial velocity it needs to go, but if we need it I guess we can come up with a formula... X*1.2? I’m not sure.

Homework Equations


I am assuming these will be used somehow:
Vx = Vi*Cos(Angle)
Vy = Vi*Sin(Angle)

The Attempt at a Solution


[/B]
Please bear with me and my absurd attempt.. I guess we can start with the easy one..
Vy2 = 2*Gravity*(Y+5) right?
We found the Vy needed to reach the maximum height.

Now to find the Vx component the angle it should be shot at, the initial velocity... can we or can we not?

I’m stuck trying to figure this out.. can anyone give me a hand?

Time if needed is the distance from origin, 0 to X multiplied by 1.2. If no time is needed for this then just ignore it.

Correct me if I’m wrong... I’m pretty sure I made more than a bunch of mistakes.
Also excuse me for typos, I’m typing from my phone.
 
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  • #2
Francis Perez said:

Homework Statement


Hello, so I’ve been trying to find some possible way to calculate this:
The initial Velocity and angle needed to launch a projectile and reach the coordinates (x,y). The maximum height the ball can reach before it starts to head down to the target value should be y+5m
The variables are only given by the initial place, being (0,0), the final place, be (X,Y), the mass of the ball, the gravity, maximum height, (Y+5). The time it will take there, I don’t know, it’s not necessary to find the time, just the angle and initial velocity it needs to go, but if we need it I guess we can come up with a formula... X*1.2? I’m not sure.

Homework Equations


I am assuming these will be used somehow:
Vx = Vi*Cos(Angle)
Vy = Vi*Sin(Angle)

The Attempt at a Solution


[/B]
Please bear with me and my absurd attempt.. I guess we can start with the easy one..
Vy2 = 2*Gravity*(Y+5) right?
We found the Vy needed to reach the maximum height.

Now to find the Vx component the angle it should be shot at, the initial velocity... can we or can we not?

I’m stuck trying to figure this out.. can anyone give me a hand?

Time if needed is the distance from origin, 0 to X multiplied by 1.2. If no time is needed for this then just ignore it.

Correct me if I’m wrong... I’m pretty sure I made more than a bunch of mistakes.
Also excuse me for typos, I’m typing from my phone.
You are on the right track. Now that you know the initial vertical velocity component, can you use it to calculate the time required to reach the target height?
 
  • #3
tnich said:
You are on the right track. Now that you know the initial vertical velocity component, can you use it to calculate the time required to reach the target height?
So let’s see.. tMaxHeight = Vy/g..
Then I find with this tFromMaxHToY =sqrt( ((Y+5)*2)/g)

Then.. tToPointY =tMH+tMHtoY

And now at this time, I have to use the X and the time to find the Vx, right? So Vx = X/t?
After that.. how do we find the optimal angle and initial velocity for these points to be reached?
 
  • #4
Francis Perez said:

Homework Statement


Hello, so I’ve been trying to find some possible way to calculate this:
The initial Velocity and angle needed to launch a projectile and reach the coordinates (x,y). The maximum height the ball can reach before it starts to head down to the target value should be y+5m
The variables are only given by the initial place, being (0,0), the final place, be (X,Y), the mass of the ball, the gravity, maximum height, (Y+5). The time it will take there, I don’t know, it’s not necessary to find the time, just the angle and initial velocity it needs to go, but if we need it I guess we can come up with a formula... X*1.2? I’m not sure.

Homework Equations


I am assuming these will be used somehow:
Vx = Vi*Cos(Angle)
Vy = Vi*Sin(Angle)

The Attempt at a Solution


[/B]
Please bear with me and my absurd attempt.. I guess we can start with the easy one..
Vy2 = 2*Gravity*(Y+5) right?
We found the Vy needed to reach the maximum height.

Now to find the Vx component the angle it should be shot at, the initial velocity... can we or can we not?

I’m stuck trying to figure this out.. can anyone give me a hand?

Time if needed is the distance from origin, 0 to X multiplied by 1.2. If no time is needed for this then just ignore it.

Correct me if I’m wrong... I’m pretty sure I made more than a bunch of mistakes.
Also excuse me for typos, I’m typing from my phone.

There are really two types of x and y at work here, and you must distinguish between them by appropriate notation, since otherwise you are likely to make mistake after mistake after mistake, and still end up with an unreadable mess. There are the x = x(t) and y = y(t) during the flight, and the final, target values. Calling the final (target) values xf and yf eliminates the difficulty.

So: do you know the formula for y = y(t), the height of the projectile at time t, in terms of mass m, initial vertical velocity Vy, time t and acceleration of gravity g? The point is that you want to know which value of t (call it t*) makes y(t) the largest, since that will be the time giving the maximum height. Then you can substitute t = t* into the formula for y(t) to get the actual maximum height y_max in terms of m, Vy, and g.

You are given the condition y_max = yf + 5, so that gives you one equation. Other given stipulations in the problem will give you other equations.
 
Last edited:
  • #5
So I’m guessing:
Vy= sqrt(2*Grav*(Yf+5))
For Y(t)Max = Yf +5
Yf+5= Vy(t) - (1/2)g*t2
Or
Yf+5 = (sqrt(2*Grav*(Yf+5)))*t - (1/2)g*t2

I’m hoping this is going to work no matter where the Target, (Xf[/SUB, Yf), are at. The projectile will do an arc 5 meters above the Yf value before going down to reach the target.
 
  • #6
Francis Perez said:
So let’s see.. tMaxHeight = Vy/g..
Then I find with this tFromMaxHToY =sqrt( ((Y+5)*2)/g)

Then.. tToPointY =tMH+tMHtoY

And now at this time, I have to use the X and the time to find the Vx, right? So Vx = X/t?
After that.. how do we find the optimal angle and initial velocity for these points to be reached?
You are on the right track, but you have made an error in your equation for ##t_{FromMaxHToY}##. How far does the projectile fall between height Y+5 and height Y?
You have the right idea about solving for ##V_x##.
As to your other question, you have written equations for ##V_x## and ##V_y## in your Relevant Equations section. How would you solve those two equations for θ and the magnitude of the velocity?
 
  • #7
tnich said:
You are on the right track, but you have made an error in your equation for ##t_{FromMaxHToY}##. How far does the projectile fall between height Y+5 and height Y?
You have the right idea about solving for ##V_x##.
As to your other question, you have written equations for ##V_x## and ##V_y## in your Relevant Equations section. How would you solve those two equations for θ and the magnitude of the velocity?

Oooh I’m dumb.. so the equation ##t_{FromMaxHToY}## would be sqrt(5*2)/g

Then.. I was thinking instead of using the angles they shoot at. I should instead find the Time when the projectile reaches Yf+5. To do this I have tried finding X-Magnitude from Xi to Xf. Then I tried using the time from max height to do find the distance at in the xplane when Y reaches its max height. TMaxHeight / Vx = XMaxHeight

Then I used arctan to find the angle it needs to shoot directing towards MaxHeight

Angle = Arctan(YMaxHeight/XMaxHeight

ProjectileAngle = (Xf-Xi) * Tan(Angle)

The initial and final velocity should be, and correct me if wrong, but I think I heard it somewhere that Projectile Velocity should be Sqrt(Vx2+Vy2) = Vi

But this only works when the projectile target is at the same level of the shooting target. I guess that when you elevate the projectile, you have to multiply the max height X coordinate and let that be the new x value that the project wants to reach, and on its way down, it will touch the actual (Xf,Yf) Target. I’m no rocket scientist but I believe it’s what I should do.. I know it’s a mess so please simplify if need be
 
  • #8
Francis Perez said:
Then I used arctan to find the angle it needs to shoot directing towards MaxHeight

Angle = Arctan(YMaxHeight/XMaxHeight
If you shoot a projectile up at an angle and directed straight at some point, you can be sure it will not hit that point but pass below it. Or have I misunderstood what you are doing here?
 
  • #9
haruspex said:
If you shoot a projectile up at an angle and directed straight at some point, you can be sure it will not hit that point but pass below it. Or have I misunderstood what you are doing here?
Yo have misunderstood me, it is not dierected exactly at the point where target (Xf,Yf) is located at. It’s actually getting the angle that is shooting to get to the Max height. Then using that angle to find the height above the Y value that the projectile must aim at in order to arc down to the (Xf,Yf) value. BUT when the target is raised. It no longer works when you set the Distance in the X-Plane to Xf, rather, what I come up with is to find the X value at the time that Y reaches the max height, then multiply that by two and let that be my new Target X. This was the projectile will try to make its way there and on its way there, pass through the elevated (Xf,Yf) target.

Please correct if any of this is wrong.
 
  • #10
Well, I was doing trial and error. It seems that I should use Vy=Vi(sin(Theta)) from the very beginning.

Vi is correct, (Vy2+Vx2)

So I did and I got Arcsin(Vy/Vi) = theta

Then I did Xf*Tan(Theta) to get the the point at which my projectile should aim and add that to the Yf.I think I was overthinking this, haha.

Did I do this right now?

EDIT: It seems to be a problem when I set the Target point below the Origin.
 
  • #11
Thank you everyone, I found the formula :)
 

Related to Find the Initial Velocity and Launch Angle for a Particular Trajectory

1. How do I find the initial velocity for a specific trajectory?

To find the initial velocity for a particular trajectory, you will need to know the distance and time of flight. Using the equation v = d/t, where v represents the initial velocity, d represents the distance, and t represents the time of flight, you can solve for the initial velocity.

2. What is the launch angle and how does it affect the trajectory?

The launch angle is the angle at which the projectile is launched from the ground. It affects the trajectory by determining the vertical and horizontal components of the initial velocity. The higher the launch angle, the higher the projectile will go and the farther it will travel horizontally. A lower launch angle will result in a shorter and flatter trajectory.

3. Can I use the same initial velocity for different launch angles?

Yes, you can use the same initial velocity for different launch angles. However, the distance and height reached by the projectile will vary depending on the launch angle. A higher launch angle will result in a longer flight path and a lower launch angle will result in a shorter flight path.

4. What factors can affect the initial velocity and launch angle?

The initial velocity can be affected by factors such as air resistance, wind speed, and the mass of the projectile. The launch angle can be affected by the angle at which the projectile is released and any external forces acting on the projectile, such as wind or gravity.

5. How can I determine the initial velocity and launch angle experimentally?

You can determine the initial velocity and launch angle experimentally by measuring the distance and time of flight for multiple trials at different launch angles. Then, using the equation v = d/t, you can calculate the initial velocity for each trial. The launch angle can be found by measuring the angle at which the projectile was launched using a protractor or using a motion tracking software to analyze the projectile's path.

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