# Trajectory - different start and end height, limited knowns

1. Oct 14, 2015

### jambraun

1. The problem statement, all variables and given/known data
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!

2. Relevant equations

3. The attempt at a solution

2. Oct 14, 2015

### CWatters

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

3. Oct 14, 2015

### 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 any one of those five values because each is determined by the other four.

4. Oct 14, 2015

### 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.

5. Oct 14, 2015

### jambraun

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!

6. Oct 14, 2015

### Mister T

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

7. Oct 15, 2015

### jambraun

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 in to 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 in to 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.

8. Oct 15, 2015

### 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,

9. Oct 15, 2015

### CWatters

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

10. Oct 15, 2015

### Mister T

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 $\theta$.

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

Last edited: Oct 15, 2015
11. Oct 15, 2015

### jambraun

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.

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.

#### Attached Files:

• ###### Y.PNG
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12. Oct 15, 2015

### Mister T

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

No.

The work you did looks correct to me.

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

Last edited: Oct 15, 2015
13. Oct 15, 2015

### jambraun

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.

14. Oct 15, 2015

### Mister T

I think you've got it!

15. Oct 16, 2015

### 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.

16. Oct 16, 2015

### 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.

Last edited: Oct 16, 2015
17. Oct 16, 2015

### 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?

18. Oct 16, 2015

### 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

19. Oct 16, 2015

### jambraun

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

20. Oct 16, 2015

### Mister T

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.

Last edited: Oct 16, 2015