Question

A body reaches the ground in $3s$ when it is released with zero velocity from the top of a building. The height of the building is:

• A. $14.7m$
• B. $24.4m$
• C. $44.1m$
• D. $66.2m$

Answer 1

Answer: (C)

$44.1m$

The given body falls from the top of a building reaches the ground in $3s$ and thus height of the building using third equation of motion is :
$S=ut+\frac{1}{2}a{t}^2$
But the ball falls from the top of the building and initial velocity is zero $u=0$).
$S=0+\frac{1}{2}a{t}^2$
$S=\frac{1}{2}9.8\times (3{)}^2=\frac{88.2}{2}=44.1$

The height of the building is $44.1m$.