Question

A stone is dropped from a cliff. Its speed after it has fallen 100mt.is:(A)9.8ms-1 (B)19.6ms-1(C)44.3ms-1 (D)98ms-1​

Answer 1

Answer:

The stone is dropped from cliff,

Therefore

u ( initial speed) = 0

V = final speed

Distance (s) = 100m

$acceleration = a = g = 9.8 \frac{m}{ {sec}^{2} } \\ \\ {v}^{2} = {u}^{2} + 2as \\ {v}^{2} = 0 + 2 \times 9.8 \times 100 \\ v = \sqrt{(2 \times 9.8 \times 100)} \\ v = 44.7 \frac{m}{sec}$