Question

A body is thrown vertically upwards from the top A of a tower. It reaches the ground in $t_1$ seconds. If it thrown vertically downwards from A with same speed it reaches the ground in $t_2$ seconds. If it is allowed to fall freely from A, then the time it takes to reach the ground is given by:

• A. $t=\frac{t_1+t_2}{2}$
• B. $t=\sqrt{\frac{t_1^2+t_2^2}{2}}$
• C. $t=\sqrt{t_1{t}_2}$
• D. $t=t_1+t_2$

Answer 1

The correct option is C $t=\sqrt{t_1{t}_2}$

From fig(A):

$-h=u{t}_1-\frac{1}{2}g{t}_1^2\Rightarrow \frac{h}{t_1}=-u+\frac{1}{2}g{t}_1..........\left(i\right)$

From fig(B):

$-h=-u{t}_2-\frac{1}{2}g{t}_2^2$
$\Rightarrow \frac{h}{t_2}=u+\frac{1}{2}g{t}_2.........\left(ii\right)$

From fig(C):
$-h=-\frac{1}{2}g{t}_3^2$
$\Rightarrow h=\frac{1}{2}g{t}_3^2.........\left(iii\right)$

From (i) and (ii):
$h[\frac{1}{t_1}+\frac{1}{t_2}]=\frac{1}{2}g(t_1+t_2)........\left(iv\right)$

$\therefore$ By eq.(iii) and (iv)
$\frac{1}{2}g{t}_3^2=\frac{1}{2}g{t}_1{t}_2$

$\Rightarrow t_3=\sqrt{t_1{t}_2}$