Question

A body is projected vertically downwards from $A$, the top of the tower reaches the ground in $t_1$ seconds. If it is projected upwards with same velocity, it reaches the ground in $t_2$ seconds. At what time will it reach the ground if it is dropped from A?

• A. $\sqrt{t_1/t_2}$
• B. $\sqrt{t_2/t_1}$
• C. $\sqrt{t_1{t}_2}$
• D. $t_1{t}_2$

Answer 1

Answer: (C)

For downward motion,

$h=u{t}_1+\frac{1}{2}g{t}_1^2...(i)$

For upward motion,

$h=-u{t}_2+\frac{1}{2}g{t}_2^2...(ii)$

From $(i)$ & $(ii)$

$u{t}_1+\frac{1}{2}g{t}_1^2=-u{t}_2+\frac{1}{2}g{t}_2^2$

$\Rightarrow u(t_1+t_2)=\frac{1}{2}g(t_2^2-t_1^2)$

$\therefore h=\frac{g}{2}(t_1-t_1)t_1+\frac{g}{2}g{t}_1^2=\frac{g{t}_2{t}_1}{2}$

$\therefore$ Required time $t=\sqrt{\frac{2h}{g}}$

$=\sqrt{t_1{t}_2}$.