Posted by **Jessica** on Tuesday, February 22, 2011 at 12:19pm.

A stone is thrown from a 50m high cliff and lands 5 seconds later, 40m from the base of the cliff. At what speed and angle was the stone thrown?

- Physics -
**drwls**, Tuesday, February 22, 2011 at 1:25pm
The horizontal velocity component must be Vx = 40/5 = 8 m/s

The initial vertical component Voy must satisfy this equation

Voy*T - (g/2)T^2 = -50

where T = 5 s, so

Voy = 14.5 m/s

Speed when thrown = sqrt(Voy^2 + Vx^2]

= sqrt[(14.5)^2 + 8^

= 16.6 m/s

The ratio Voy/Vx is the tangent of the launch angle

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