Question

A ball is dropped from the top of a tower of the height of $100m$. Simultaneously, another ball was thrown upward from the bottom of the tower with a speed of $25m/s$ ($g=10m/s^2$ ). These two balls would cross each other after a time:

• A. $1s$
• B. $2s$
• C. $3s$
• D. $4s$

Answer 1

The correct option is D

$4s$

Step 1: Given Data

Height of the tower $h=100m$

Acceleration due to gravity $g=10m/s^2$

Let the ball dropped from the top of the tower be $A$ and the ball thrown upwards from the bottom of the tower be $B$.

The initial velocity of ball $A$, $u=0$

The initial velocity of ball $B$, $u=25m/s$

Let the height from the top of the tower to the point of intersection be $h_1$.

Let the height from the bottom of the tower to the point of intersection be $h_2$.

Let the time taken for the intersection of the two balls be $t$.

Step 2: Calculate the height $h_1$

According to Newton's second equation of motion we get,

$S=ut+\frac{1}{2}a{t}^2$

$\Rightarrow h_1=0+\frac{1}{2}\times 10\times t^2$

$\Rightarrow h_1=5t^2$ $\left(1\right)$

Step 3: Calculate the height $h_2$

Since the motion of ball $B$ is against the gravity according to Newton's second equation of motion we get,

$S=ut+\frac{1}{2}a{t}^2$

$\Rightarrow h_2=25t-\frac{1}{2}\times 10\times t^2$

$\Rightarrow h_2=25t-5t^2$ $\left(2\right)$

Step 4: Calculate the time

Upon adding equation $\left(1\right)$ and $\left(2\right)$ we get

$h_1+h_2=5t^2+25t-5t^2$

$\Rightarrow h_1+h_2=25t$

$\Rightarrow 100=25t$

$\Rightarrow t=\frac{100}{25}=4s$

Hence, the correct answer is option (D).