Exact question: A hobby rocket reaches a height of 35m and lands 259m from the launch point. What was the angle of launch?
Xmax = 259m
Ymax = 35m
Xo and Xmax are at equal y values, or heights, 0 in my chosen coordinate system
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)
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.