Physics
posted by Jessica .
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?

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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