Question

A particle is released from rest from a height of 100m. find the distance it falls through in 2,4sec g=9.8ms^-2

Answer 1

Initial velocity = 0 m/s

Height from which it is released = 100 m

a) time (t) = 2 seconds

a = 9.8 m/s^2

Let the distance covered be s.

$s = ut + \frac{1}{2} \times a \times t^2$

$s = 0 + \frac{1}{2}\times 9.8 \times 2 \times 2$

s = 19.6 m

b) t = 4 seconds

Let the distance covered be y

$y = ut + \frac{1}{2} \times a \times t^2$

$y = 0 + \frac{1}{2}\times 9.8 \times 4 \times 4$

y = 78.4 meters