Question

2. A pebble is dropped off a 300 m tall cliff. Using the conservation of energy
equation, find out how fast is it going right before it hits the ground?​

Answer 1

Answer:

77.45 m/Sec

KE before landing = PE at leaving

1/2 m vsquare = mgh

v square= 2 g H

v = Root( 2x10x300)

v= Root( 6000)

V = 77.45 m/sec