Trajectory - different start and end height, limited knowns

[jambraun]

Homework Statement

A projectile is fired from a cannon at X-distance of 25m (X=25), height 10m (y=10), and fixed angle (53-degrees). After 3 seconds, the projectile hits its mark at different height 50m (y=50), and at a range of 175m (X=200). Gravity is normal and there is no wind resistance.

What is the original velocity of the projectile, required equations to find the original velocity, and the equation showing the position of the projectile at any value of 'Time'.

Gravity = -9.8m/s
Starting X = 25m
Range (X to target) = 175m
Firing Angle = 53-degrees
Starting height (Y) = 10m
Target/End height (Y) = 50m
Time in air = 3s

Unknowns = Velocity(original X), Velocity (original Y)

I'm looking to apply this to a game I'm programming but haven't been able to make headway on the equations for months now. I've come close with an example from Khan (https://www.khanacademy.org/science...launching-and-landing-on-different-elevations) and a few others but I can't seem to re-write the equations to solve for initial velocity, and then the proper equation to find the position of the projectile at any value of 't' (time), but I think that's a standard equation once I find the original velocity. Any help is much appreciated!

Homework Equations

The Attempt at a Solution

 

[CWatters]

Show your attempt so we can see where you are going wrong. Forum rule.

 

[Mister T]

Your situation violates the laws of physics. The laws of physics establish a relationship between the free fall acceleration, height, range, time, and launch angle. The numbers you state don't fit that relationship. You have to abandon anyone of those five values because each is determined by the other four.

 

[jambraun]

Thanks CWatters, my major notes are in my notebook at home; I'll post them in the morning. I think my primary issue is displacement of X to compensate for the differing heights. I have no problem calculating the launch and land from the same altitude which is pretty straight forward.

 

[jambraun]

Your situation violates the laws of physics. The laws of physics establish a relationship between the free fall acceleration, height, range, time, and launch angle. The numbers you state don't fit that relationship. You have to abandon anyone of those five values because each is determined by the other four.

I knew something was up...I can be flexible on time or gravity. let me post the equations I've been working with in the morning but I still think my hangup is with the height displacement. Thanks!

 

[Mister T]

Change the time to 6.26 seconds and then you can solve for $v_o$.

 

[jambraun]

Change the time to 6.26 seconds and then you can solve for $v_o$.

6.26s is too long (unfortunately). Can I fudge gravity or angle to decrease the time? I still need to figure out how to displace the distance so the target is struck at the right height.

We'll still have to talk about how to account for displacement I think. Here's what I was trying to use:

3 step process and 2 primary equations for determining initial velocity:

Primary Equations
X=v*t*cos(theta)
Y=v*t*cos(theta) - .5*g*t^2

Step 1 (create a substitution for Y from the equation for X):
Initial Velocity of X * Time = vt = x/cos(theta)

Step 2 (plug equations from X into Y in order to isolate t):
Isolating time from the equation for Y
Y=v*t*cos(theta) - .5*g*t^2
== Y = VxTx - .5gt^2
=== Y-(VxTx)=-.5gt^2
== t = SQRT((2(Y-(VxTx)))/g)
== t = (SQRT(2TxVx-2y))/SQRT(g)

Step 3 (plug t back into X in order to find velocity)
X=v*t*cos(theta)

See my notes below from scan. When I used this in a graphing calculator it appeared to come out ok, but it doesn't account for height displacement.

 

[CWatters]

I haven't looked at your latest post yet but the approach I would take is to subtract 10m from both the launch and target heights to simplify the problem. It won't effect the launch angle, speed, time of flight etc,

 

[CWatters]

X=v*t*cos(theta)
Y=v*t*cos(theta) - .5*g*t^2

The eqn for y is incorrect. Should be sin(theta) not cos(theta).

 

[Mister T]

6.26s is too long (unfortunately). Can I fudge gravity or angle to decrease the time?

Yup. Either one or both.

Step 1 (create a substitution for Y from the equation for X):
Initial Velocity of X * Time = vt = x/cos(theta)

You have

$v_ot=\displaystyle \frac{\Delta x}{\cos \theta}$.

Just solve for $v_o$.

$v_o=\displaystyle \frac{\Delta x}{t \cos \theta}$.

Plug in the numbers and you're done!

Of course, that won't give you the desired rise of 40 m unless you change $g$. Other options are choosing different values for $t$, or $\theta$.

$\Delta y=(v_o \sin \theta) t - \frac{1}{2}gt^2$.

 

[jambraun]

Yup. Either one or both.
You have

$v_ot=\displaystyle \frac{\Delta x}{\cos \theta}$.

Just solve for $v_o$.

$v_o=\displaystyle \frac{\Delta x}{t \cos \theta}$.

Plug in the numbers and you're done!

Of course, that won't give you the desired rise of 40 m unless you change $g$. Other options are choosing different values for $t$, or $\theta$.

$\Delta y=(v_o \sin \theta) t - \frac{1}{2}gt^2$.

Sorry, the equation variables aren't something I can guess at while running for the program to work. Plus, doesn't calculating the initial velocity from the original equation of X always land short of the mark? I believe that's why it's a 3 step process, obtaining the true time in the air from the Y.

The eqn for y is incorrect. Should be sin(theta) not cos(theta).

How's this look? More importantly, am I headed in the right direction? Besides the math being correct, is it the correct math to use? I'd be up for using different equations if you know of the correct ones to use to project a flight path that MUST hit a point in space with a few constraints.

 

[Mister T]

Sorry, the equation variables aren't something I can guess at while running for the program to work.

I don't understand this comment. I was trying to help you answer your question. You wanted to know the launch speed.

$v_o=\displaystyle \frac{\Delta x}{t \cos \theta}$.

Plug in the numbers and you're done!

The numbers you gave us were t = 3 s, Δx = 175 m, and θ = 53°.

Plus, doesn't calculating the initial velocity from the original equation of X always land short of the mark?

No.

How's this look?

The work you did looks correct to me.

More importantly, am I headed in the right direction?

Well, let's see. You wanted to alter the value of g so that you could preserve the shorter time of 3 seconds and the rise of 40 m. I showed you how to do that in Post #10. You simply plug all the numbers into that last equation and solve it for g.

$\Delta y=(v_o \sin \theta) t - \frac{1}{2}gt^2$.

Or, you could solve the equation you came up with for g. Either way, you get the same answer. Or you could alter any of the other variables to your liking, provided they are solutions to your equations

 

[jambraun]

I don't understand this comment. I was trying to help you answer your question. You wanted to know the launch speed.

Oh, I got lost when you said I could change the gravity to get the needed rise. I didn't understand if I'm solving for gravity or just choosing a greater gravity? Also, for solving the difference in Y, are you saying:

But I don't have to use steps 1 and 2 from notes, just plug in the Delta-Y of 40, the found velocity from X, and expected time of 3?

Thanks for the help, btw. I'll try graphing it out in the morning to check the curve.

 

[Mister T]

I think you've got it!

 

[jambraun]

ok, I need someone to check my math... here's what happens when I plug in the variables:

The hardest equation by far is to find gravity (which must be incorrect).

Alternatively, if I solve for gravity by hand...
40 = (175/(2*cos(70)))*2*sin(70)-.5*g*2^2
40 = 212.52-2g...
Help me out where to go from here? I know this answer is MUCH different than the auto-solve.

 

[CWatters]

You have two equations (one for x and one for y). I would substitute to eliminate "t" and rearrange the result to give an equation for g. Then plug in the values for x and y for the target and v.

Correction:

Ok looking back I see you have defined t but not V. So best substitute to eliminate V rather than t and rearrange the result to give an equation for g. Then plug in the values for x and y for the target and t.

 

[jambraun]

I'm failing in the substitution and rearrange steps. It's always been my weakness.

Can I beg you to show me what the correct equation for g is?

 

[CWatters]

Unfortunately forum rules discourage us from just giving you the answer. If you show us your attempt/working we can point out you where you went wrong

 

[jambraun]

Alternatively, if I solve for gravity by hand...
40 = (175/(2*cos(70)))*2*sin(70)-.5*g*2^2
40 = 212.52-2g...
Help me out where to go from here? I know this answer is MUCH different than the auto-solve.

I thought I did :/ the answer to the above came out to be -(40/106) which is still waaaay off the mark.

I'm posting from my phone so I can't do anything elaborate for a bit

 

[Mister T]

Alternatively, if I solve for gravity by hand...
40 = (175/(2*cos(70)))*2*sin(70)-.5*g*2^2
40 = 212.52-2g...

Your calculator appears to be in radian mode. In your program the computer is treating the angles as if they're expressed in radians, not degrees, so that's giving you an incorrect value for $v_o$, and messing up the solver results and the plot, too.

Somehow, though, the value for $g$ of 220 m/s² is correct.

By the way, when you change t from 3 s to 2 s, and θ from 53° to 70° it makes the projectile hit the target on the way up instead of on the way down. In fact, if the target doesn't alter the projectile's path it'll take 20 seconds for it to reach the crest! In your Post #1 sketch you show the projectile hitting the target on the way down.

 

[CWatters]

I seem to be getting different answers.

The equation I got for g was..

g = 2(xtan(θ) - y) / t²

Using... .
x = 200-25 = 175m
y = 50-10 = 40m
t = 3
angle = 70
I get g about 98m/s/s

Using...
x = 200-25 = 175m
y = 50-10 = 40m
t = 2
angle = 53
I get g about 96m/s/s

 

[jambraun]

Sorry! I changed my values after you mentioned I could alter some variables to have the time change...that was probably a bad idea since it changes the help I'm asking for. Between you and Cwatters I'll write up my final equations and repost with the completed version. I think the answer is in here!

 

[Mister T]

I seem to be getting different answers.

The equation I got for g was..

g = 2(xtan(θ) - y) / t²

Using... .
x = 200-25 = 175m
y = 50-10 = 40m
t = 3
angle = 70
I get g about 98m/s/s

Using...
x = 200-25 = 175m
y = 50-10 = 40m
t = 2
angle = 53
I get g about 96m/s/s

You have different combinations than the OP. He had the angle of 70° for the time of 2 s. And the angle of 53° for the time of 3 s.

 

[CWatters]

Ah ok I used wrong combination of values. If I use..

x = 200-25 = 175m
y = 50-10 = 40m
t = 2
angle = 70
I do get g about 220m/s/s

 

[jambraun]

I think that may have done it! Thanks for the help. I'll drop this into my program and see if it hits the target :)

 

[Mister T]

Just so long as you understand that your launch angle is 70 radians, or about 4010°.

And as I said before the projectile hits the target while rising. So the trajectory, while actually being a piece of a parabola, might look more like a straight line.

 

[jambraun]

lol, you're absolutely right. Here's the conversion using degrees. I'll have to play with it so the projectile stays on the playfield / looks believable:

 

[Mister T]

All you have to do is go back to the values you were using up until yesterday: 53° and 3 seconds.