Question

A stone dropped from the roof of a building takes 4 second of to reach the ground the calculate the height of the building?

Answer 1

initial velocity = 0
time taken = 4s
acceleration = g = 9.8m/s^2

using 2nd equation of motion
s= ut +(1/2)at^2
= (0)(4) +(1/2)(9.8)(4)(4)
= 78.4


So,
distance covered by that stone was 78.4m

Answer 2

Given:

time taken $=4 seconds$

initial velocity $= 0$

acceleration $= 9.8 m/sec^{2}$

To proof:

We need to find the height of the building

Explanation:

• We can find the height of the building by simply calculating the equation of the motion.

SOLUTION:

$height = ut + \frac{1}{2} a t^{2} \\\\= 0 *4 + \frac{1}{2} * 9.8 * 4 *4\\\\= \frac{1}{2}*9.8*16\\\\= 78.4m$

Hence, the distance covered by the stone is $78.4m$