Question

Two balls are dropped from different heights at different instants. Second ball is dropped $2\text{s}$ after the first ball. If both the balls reach the ground simultaneously, after $5\text{s}$ of dropping the first ball, then the difference between the initial heights of the two balls will be
$(\text{Take}g=9.8{\text{m/s}}^2)$

• A. $58.8\text{m}$
• B. $78.4\text{m}$
• C. $98.0\text{m}$
• D. $117.6\text{m}$

Answer 1

The correct option is B $78.4\text{m}$


Since, both ball reach ground after $5\text{sec}$ dropping $1^{st}$ ball, so the time taken by the $1^{st}$ ball to reach the ground is $t_1=5\text{s}$.

From diagram, height travelled by $1^{st}$ ball in $5\text{s}$ is $h_1.$

So, using equation of motion for $1^{st}$ ball, we get

$h_1=\frac{1}{2}g{t}_1^2$

$\Rightarrow h_1=\frac{1}{2}g(5{)}^2=\frac{25}{2}g$

Since the $2^{nd}$ ball is dropped $2\text{s}$ after the $1^{st}$ ball, so the time taken by it to reach the ground is $(t_2=5-2=3\text{s})$.

Similarly, height covered by $2^{nd}$ ball in $3\text{s}$ is $h_2$

$h_2=\frac{1}{2}g{\left(3\right)}^2=\frac{9}{2}g$

So, the difference in the heights,

$h_1-h_2=\frac{25}{2}g-\frac{9}{2}g$

$\Rightarrow h_1-h_2=8g$

$\Rightarrow h_1-h_2=8\times 9.8=78.4\text{m}$

Hence, option $\left(b\right)$ is correct.