A woman stands on a cliff 20 m high and throws a rock with speed 20 m/s at an angle of 30ø with the horizontal

Question

the magnitude of the velocity with which the rock hits the ground below is

Answer 1

the magnitude is found by looking at conservation of energy

kinetic energy at the time of release plus the conversion of potential energy (gravity) equals the kinetic energy at impact (since air friction is ignored)

.5*m*v^2=g*m*20+.5*m*20*20

v^2=40*g+400
using g=10

v=sqrt(800)
=28.3 m/s

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