Question

A particle is dropped from a height $h=100\text{m}$, from the surface of a planet. If in the last $\frac{1}{2}\text{sec}$ of its journey, it covers $19\text{m}$, the value of acceleration due to gravity on that planet is:

• A. $8{\text{m/s}}^2$
• B. $10{\text{m/s}}^2$
• C. $6{\text{m/s}}^2$
• D. $4{\text{m/s}}^2$

Answer 1

The correct option is A $8{\text{m/s}}^2$


As we know,
Area under the velocity vs time graph gives displacement.

Area of shaded trapezium
$=\frac{1}{2}{v}_2\left(t\right)-\frac{1}{2}{v}_1(t-\frac{1}{2})=\text{displacement in last 1/2 sec}$
Here,
$v_1=g(t-\frac{1}{2})$ , $v_2=gt$
$\Rightarrow \frac{1}{2}g\left(t^2\right)-\frac{1}{2}g(t-\frac{1}{2}{)}^2=19....(1)$

Distance travelled in $t\text{sec}$
$\frac{1}{2}g{t}^2=100$ [total height]
$t=\sqrt{\frac{200}{g}}....\left(2\right)$

From eqn (1) and (2)
$g\left[\frac{4t-1}{8}\right]=19$
$\frac{152}{g}=[4\times \sqrt{\frac{200}{g}}-1]$
or $g=8{\text{m/s}}^2$