Question

A body is thrown vertically upwards from $A$, the top of the tower. It reaches the ground in time $t_1$. If it is thrown vertically downwards from $A$ with the same speed, it reaches the ground in time $t_2$. 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 $t=\sqrt{t_1{t}_2}$

Let the body is projected vertically upwards from $A$ with a speed $u_0$.

Using equation, $s=ut+\frac{1}{2}a{t}^2$

For case $\left(1\right)-h=u_0{t}_1-\frac{1}{2}g{t}_1^2.....\left(1\right)$

For case $\left(2\right)-h=u_0{t}_2-\frac{1}{2}g{t}_1^2....\left(2\right)$

Subtracting eq $\left(2\right)$ from $\left(1\right)$, we get

$0=u_0(t_2+t_1)+\frac{1}{2}g(t_2^2-t_1^2)$

$\Rightarrow u_0=\frac{1}{2}g(t_1-t_2).....\left(3\right)$

Putting the value of $u_0$ in eq $\left(2\right)$, we get

$-=h=-\left(\frac{1}{2}\right)g(t_1-t_2)t_2-\left(\frac{1}{2}\right)g{t}_2^2$

$\Rightarrow h=\frac{1}{2}g\left(t_1{t}_2\right).....\left(4\right)$

For case $3,u_0=0,t=?$

$-h=0\times t-\left(\frac{1}{2}\right)g{t}^2$

$h=\left(\frac{1}{2}\right)g{t}^2$

Comparing eq. $\left(4\right)$ and $\left(5\right)$, we get

$\frac{1}{2}g{t}^2=\frac{1}{2}g{t}_1{t}_2\therefore t=\sqrt{t_1{t}_2}$.