Question

A ball thrown vertically upwards at a certain speed from the top of a tower of height $50\text{m}$ m reaches the ground after $9\text{seconds}$. Another ball thrown vertically downwards with same speed from the same tower reaches the ground in $4$ seconds. How much time will the ball take if it is just dropped from the tower?

• A. $6s$
• B. $7s$
• C. $5s$
• D. $8s$

Answer 1

The correct option is A $6s$

Taking upward direction as (+ve) and downward direction as (-ve) y-axis.


For the ball in the first case,
$s=-h,a=-g,t=t_1,u=u_0$
Applying the equation of motion for first case, we get,
$-h=u_0{t}_1+\frac{1}{2}(-g)t_1^2$

Multiplying this whole equation by $t_2$, we get,
$-h{t}_2=u_0{t}_1{t}_2+\frac{1}{2}(-g)t_1^2{t}_2...\left(i\right)$


For the ball in the second case,
$s=-h,a=-g,t=t_2,u=-u_0$

.
Simillarly for second case,
$-h=(-u_0)t_2+\frac{1}{2}(-g)t_2^2$
Multiplying this whole equation by $t_1$, we get,
$-h{t}_1=(-u_0)t_2{t}_1+\frac{1}{2}(-g)t_2^2{t}_1....\left(ii\right)$

When the ball is just dropped,
$-h=\frac{1}{2}(-g)t^2$
$\Rightarrow h=\frac{1}{2}\left(g\right)t^2....\left(iii\right)$

Adding the equations $\left(i\right)$ and $\left(ii\right)$, we get,
$-h(t_1+t_2)=-\frac{1}{2}g{t}_1{t}_2(t_1+t_2)$
$\Rightarrow h(t_1+t_2)=\frac{1}{2}g{t}_1{t}_2(t_1+t_2)$
$\Rightarrow h=\frac{1}{2}g{t}_1{t}_2...\left(iv\right)$

Comparing equations $\left(iii\right)$ and $\left(iv\right)$, we get,
$\frac{1}{2}g{t}^2=\frac{1}{2}g{t}_1{t}_2$
$t=\sqrt{t_1{t}_2}$
$t=\sqrt{9\times 4}=\sqrt{36}=6s$