Finding Lanch Angle (without initial velocity)

peterbishop

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

Exact question: A hobby rocket reaches a height of 35m and lands 259m from the launch point. What was the angle of launch?

We know...
Xmax = 259m
Ymax = 35m
Xo and Xmax are at equal y values, or heights, 0 in my chosen coordinate system

2. Relevant equations

1) Xmax = (Vo^2 * sin(2 * theta))/g
2) Ymax = (Vo * sin(theta))^2 / g
3) Vy at max height is zero
4) time of Xmax = (2Vo * sin(theta)) / g
5) time of Ymax = (Vo * sin(theta)) / g
6) Y = Yo + Vosin(theta)t - (1/2)gt^2
7) X = Xo + Vocos(theta)t
8) Vfy^2 - Voy^2 = -2gh (or something like that, I haven't used this equation in conjunction with projectile motion before)

3. The attempt at a solution

This problem has stumped me for a long time. I remember I had something like it in high school as well that I had a hard time with. It would take up way too much room to type out all my wrong solutions, but I'll go over some of the stuff I've tried today. I tried using equations 1 and 2, setting them equal to their respective known values. I tried to solve for Vo in the Ymax equation, getting Vo = 26.2/sin(theta). I tried plugging that into the Xmax equasion, getting a messy kind of simplified 3.7 = sin(2*theta)/(sin^2(theta)). I wasn't sure what to do with that then, trig is not my strong point and there has to be a better way to do this problem. I've tried using equations 6 and 7, solving for time in 7 and plugging it back into 6 but that didn't work either. Any advice on how to go about solving this problem and ones like it would be helpful, I'm sure it's just something I'm overlooking or a substitution I haven't thought about.

ehild

Your second equation is wrong: The correct one is Ymax = (Vo * sin(theta))^2 /(2g).

(The vertical motion of the projectile is described with the equations (6) and v_y = vo sin(theta)-gt.
As you know correctly, v_y=0 at maximum height, the time to reach it is t = vo sin(theta)/g. Substitute for t in eq. 6, you get y(max)=(vo sin(theta))^2/(2g) )

Knowing y(max), you get vo sin(theta). The first equation yields vo^2 * sin(2 * theta). You know the trigonometric eq.
sin(2* theta)=2 sin (theta)*cos(theta).

You have two equations for vo and theta. Cancel vo, solve for theta, then get vo.

ehild

JaredJames

http://www.physicsforums.com/showthread.php?t=434921

I'm going to point you here as well, I just answered this and it's a similar question. The only difference is you'd have to resolve for the ascent time as well.

If you can understand what I've done there, you can work out how to solve your question.

Jared

rl.bhat

Since you want only the angle of launch, find the expressions for Ymax and Xmax.
If t is he time to reach Ymax, 2t will be the time to reach Xmax.

2vo^2*sinθ*cosθ = Xmax ....(1)
vo^2*sin^2(θ) = 2*g*Ymax .......(2)

Divide (2) by (1) and solve for θ.