Question

A ball dropped from the top of the tower falls first half height of the tower in 10s. The total time spent by a ball in the air is $\left[Takeg=10m/s^2\right]$

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

Answer 1

The correct option is A $14.14s$

Let the height of tower is h. Now considering the equation as

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

$h=ut+\frac{1}{2}a{t}^2...........\left(1\right)$

$here,h=\frac{h}{2}$

In 10 sec, object is reaching half the height of tower.

Initially the body is rest so u = 0

It is also given that a = 10 ms^-2

Time to reach first half height of tower t = 10 sec

Put the values in (1)

$\frac{h}{2}=0+\frac{1}{2}\times 10\times {10}^2$

$h=1000m$

The total time spent by a ball in the air is given by the relation as

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

$1000=\frac{1}{2}\times 10\times t^2$

$t=10\sqrt{2}$

$t=14.14\sec$