Question

A body falls from the top of a building and reaches the ground $2.5s$ later. How high is the building? (Take $g=10m{s}^{-2}$)

• A. 30.6 m
• B. 31.25 m
• C. 30 m
• D. 25 m

Answer 1

The correct option is B 31.25 m

Height of the building is given by 2nd equation of motion,
$S=ut+\frac{1}{2}g{t}^2$
now since initial velocity is zero so 2nd equation of motion, become,
$S=\frac{1}{2}g{t}^2=\frac{1}{2}\times 10\times {2.5}^2=31.25m$