A hockey puck truck at ice level just clears the top of a glass wall that is 2.80 m high. The flight time to this point is 0.650 s, and the horizontal distance is 12.0 m. Find a.) the initial speed of the puck and b.) the maximum height it will reach.

1 answer

horrizontal distance=vi*cosTheta*time
vi*costheta=12/.650

now, for time in the air

hf=hi+1/2 g t^2+ vi*sinTheta*t
2.8= -4.9 t^2+vi*sinTheta*t
you know t, for vi put in 12/(cosTheta*.65)
2.8= -4.9 t^2+ sinTheta*t*12/(cosTheta*.65)

now you have an equation you can solve for Tan Theta, then theta, the launch angle.
Then, you can go back and solve for vi, and hf max (when vvertical is zero).