posted by Mbali .
A stone is thrown from a 50m high cliff and lands 4 seconds later, 40m from the base of the cliff. At what speed and angle was the stone thrown?
The time to fall, T = 4s, tells you the vertical initial velocity Vy0
-50 = Vy0 T - (1/2) gT^2
-50 = Vy0*4 - 4.9* 16 Vy0 - 78.4
4 Vy0 = 28.4
Vy0 = 7.1 m/s (upwards)
The "0" at the end of the subscript means "initial value"
For Vx, solve this equation that uses the horizontal coordinate where it hit the ground:
40 = Vx * T
Vx = 10 m/s (Vx remains the same with time, so it doesn't need a 0 subscript
The speed is sqrt[Vx^2 + Vy0^2]
and the launch angle is arctan Vy0/Vx