Question

A ball dropped from the top of tower falls first half height of tower in $10s$. The total time spend by ball in air is [Take $g=10m/s^2$ ]

• A. 14.14 s
• B. 15.25 s
• C. 12.36 s
• D. 17.36 s

Answer 1

Answer: (A)

$\frac{-H}{2}=ut-\frac{1}{2}g\times 10^2$
$\Rightarrow H=g\times 10^2$
$\Rightarrow -H=-\frac{1}{2}g{t}^2$ (Full journey)
$g\times 10^2=\frac{1}{2}g{t}^2$
$\Rightarrow t^2=200$
$\Rightarrow t=10\sqrt{2}s$
$\Rightarrow t=10\times 1.414s$
$=14.14s=t$