Question

A ball is dropped from a height. If it takes $0.20s$ to cover the last $6.00m$ before hitting the ground. Find the height from which it was dropped. Take $g=10m/s^2$.

• A. $48m$
• B. $36m$
• C. $108m$
• D. $56m$

Answer 1

The correct option is A $48m$

Uniformly accelerated motion.

So for last 6m distance travelled

As we know

For last 6 m

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

Given :-

$S=6m$

$u=?$

$t=0.2sec$

$a=g=9.8m/s^2$

$s=ut+\frac{1}{2}a{t}^2\to 6=u\left(0.2\right)+4.9\times 0.04$

$u=29m/s$

For distance $x,u=0,v=29m/s$

Now for time before last 6 m

$a=g=9.8m/s^2$

$s=\frac{v^2-u^2}{2a}\Rightarrow \frac{{29}^2-0^2}{2\times 9.8}\Rightarrow 42.05$

Total distance $=42.05+6$

$=48m$