Question

A boy releases a ball from the top of a building. it will clear the window 2m high at a distance 10m below the top in nearly.(take $g=10\raisebox{1ex}{$m$}\!\left/ \!\raisebox{-1ex}{$s^2$}\right.$)

• A. 1s
• B. 1.3s
• C. 0.6s
• D. 0.13s

Answer 1

The correct option is D

0.13s

Step 1: Given:

1. A boy releases a ball from the top of a building, and the initial velocity (u) will be zero.
2. The height of the window is 2 m.
3. Distance is 10 m below the top.
4. $g=10m/s^2$

Step 2: From the second equation of motion,

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

Where u is initial velocity, a is a constant acceleration, t is time, and s is displacement.

As u = 0,

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

For the given situation,

s = height, a = g

$\mathrm{h}=\frac{1}{2}{\mathrm{gt}}^2$

Therefore time taken,

$\mathrm{t}=\sqrt{\frac{2\mathrm{h}}{\mathrm{g}}}$

Using the above formula,

Step 3: Finding time taken,

${\mathrm{t}}_1=\sqrt{\frac{2\times 12}{10}}$

Similarly,

${\mathrm{t}}_2=\sqrt{\frac{2\times 10}{10}}$

Therefore according to the question,

$t_1-t_2=\left(1.549-1.414\right)=0.135sec$

Therefore the time taken by the ball is 0.135 seconds.

Hence, the correct option is (D).