Question

A ball of mass $100\text{g}$ is projected vertically upwards from the ground with a velocity of $98\text{m/s}$. At the same time, another identical ball is dropped from a height of $98\text{m}$ to fall freely along the same path as that followed by the first ball. After some time, the two balls collide and stick together and finally fall to the ground. The time of flight of the masses is

• A. $7.75\text{s}$
• B. $6.53\text{s}$
• C. $8.90\text{s}$
• D. $9.23\text{s}$

Answer 1

The correct option is B $6.53\text{s}$

Given, two balls are identical, So masses of both the balls are same.

Position of centre of mass of the two ball system from the ground,

$h_{cm}=\frac{(m\times 0)+(m\times 98)}{m+m}=49\text{m}$

Acceleration of the centre of mass,

$a_{cm}=\frac{(m\times g)+(m\times g)}{m+m}=9.8{\text{m/s}}^2$ (downward)

Initial velocity of the centre of mass,

$u_{cm}=\frac{(m\times 49)+(m\times 0)}{m+m}=24.5\text{m/s}$ (upward)

Let centre of mass of the system strikes the ground after time $t$.

Since, acceleration is constant, we use kinematic relation $s=ut+\frac{1}{2}a{t}^2$

$\therefore s=u_{cm}t+\frac{1}{2}{a}_{cm}{t}^2$

Substituting the data we get,

$-49=24.5t-\frac{1}{2}\times 9.8\times t^2$

$\Rightarrow -49=24.5t-4.9t^2$

$\Rightarrow t^2-5t-10=0$

Solving the quadratic equation,

$t=\frac{5\pm \sqrt{25+40}}{2}\Rightarrow t=6.53,-1.53$

Time cannot be negative.

Thus, the time of flight of the masses is $6.53\text{s}$.

Hence, option (b) is the correct answer.