Question

A body is thrown vertically upward from the top $A$ of tower. It reaches the ground in $t_{1}s$. If it is thrown vertically downwards from $A$ with the same speed it reaches the ground in $t_{2}s$. 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. $t=\frac{t_1-t_2}{2}$
• C. $t=\sqrt{t_1{t}_2}$
• D. $t=\sqrt{\frac{t_1}{t_2}}$

Answer 1

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

Taking downward direction as positive direction.
Case 1: Ball is thrown upwards with velocity $u$.
Using second equation of motion we get
$+h=-u{t}_1+\frac{1}{2}g{t}_1^2\dots \dots \left(i\right)$
Case 2: Ball is thrown downwards with velocity $u$.
Using second equation of motion we get
$+h=u{t}_2+\frac{1}{2}g{t}_2^2\dots \dots \left(ii\right)$
Multiplying Eq. $\left(i\right)$ by $t_2$ and Eq. $\left(ii\right)$ by $t_1$ and adding. We get
$h(t_1+t_2)=\frac{1}{2}g{t}_1{t}_2(t_1+t_2)$
or $h=\frac{1}{2}g{t}_1{t}_2\dots \dots \left(iii\right)$
In case of free fall, we can write
$h=\frac{1}{2}g{t}^2\dots \dots \left(iv\right)$
$\therefore t^2=t_1{t}_2$
$t=\sqrt{t_1{t}_2}$