Posted by **Mbali** on Thursday, February 26, 2009 at 12:03pm.

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?

- Physical Sciences -
**drwls**, Thursday, February 26, 2009 at 3:47pm
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

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