Question

A stone is thrown vertically upward with a speed of 10.0 ms-1 from the edge of a cliff 65m high. What will be its speed just before hitting the bottom ?
1) 3.14 m/s 2) 37.14 m/s 3) 13.71 m/s 4) 14.71 m/s

Answer 1

Answer:

V^2 - u^2 = 2aS

v^2 = u^2 + 2gH

v^2 = 10^2 + 2*9.8*65

v^2 = 1374

v = sqrt(1374) m/s

It’s speed just before hitting the bottom = sqrt(1374) m/s

Answer 2

Answer:

This answer wascorrect