Question

An aeroplane is rising vertically with acceleration $f$. Two stones are dropped from it at an interval of time $t$. The distance between them at time $t^{\prime }$ after the second stone is dropped will be

• A. $\frac{1}{2}(g+f)t{t}^{\prime }$
• B. $\frac{1}{2}(g+f)(t+2t^{\prime })t$
• C. $\frac{1}{2}(g+f)(t-t^{\prime }{)}^2$
• D. $\frac{1}{2}(g+f)(t+t^{\prime }{)}^2$

Answer 1

The correct option is B $\frac{1}{2}(g+f)(t+2t^{\prime })t$

The displacement between first stone and aeroplane after $t$ second $\left(h_1\right)=\frac{1}{2}(g+f)t^2$
After time $t$,
Velocity of aeroplane $=u+ft$
Velocity of first stone $=u-gt$
Where $u$ is velocity of aeroplane when first stone is dropped.
The relative speed of second stone with respect to first stone $=(u+ft)-(u-gt)$
$=(g+f)t$
The relative displacement between first and second stone after time $t^{\prime }\left(h_2\right)$
$=(g+f)t{t}^{\prime }$
$h_1+h_2=\frac{1}{2}(g+f)t^2+(g+f)t{t}^{\prime }=\frac{1}{2}(g+f)(t+2t^{\prime })t$