# Projectile Motion: A boy throw a ball above a barn roof

1. Feb 5, 2012

### archcos

1. The problem statement, all variables and given/known data
A boy attempts to throw a ball over the roof of a barn 11m high with an initial velocity of vsub0=12m/s. Determine the angle theta, at which the ball must be thrown so that it reaches its maximum height, which is at the roof of the barn. also find x where the boy must stand to make a throw, if the ball is thrown 1m from the ground.

2. Relevant equations
x=vsub0*costheta*time
y=Vsub0sinetheta*time-1/2 gt^2

or if max height is given 11m maybe, h=vsub0^2sin^2theta / 2g

3. The attempt at a solution

if i use equation for max height I cant get the angle but im thinking of combining the vertical and horizontal comp of displacement if I could get some hints to solving the angle or maybe the total x distance, but im stuck with a time missing then made me confused....

Last edited by a moderator: Feb 5, 2012
2. Feb 5, 2012

### vela

Staff Emeritus
Why can't you find the angle using the max-height formula? It should work.

3. Feb 5, 2012

### archcos

syntax error for me...i cant get the angle

4. Feb 5, 2012

### vela

Staff Emeritus
Why? Show your work so we can see where you're getting stuck.

5. Feb 5, 2012

### archcos

i get errors here
11=12^2sin^2theta / 2*9.8
11*2*9.8=144sin^2theta
√215.6/144=√sin^2theta
theta=arcsin(1.22) then error

6. Feb 5, 2012

### vela

Staff Emeritus
That result (no real solution) indicates it's impossible for the boy to throw the ball over the roof of the barn.

The maximum height the ball could reach occurs if he were to throw it straight up. Try calculating what that is. If it's at least 11 m, there's no possible way for the ball to make it over the barn with that initial velocity.