Q89 / 120 · VITEEE 2013 · Physics
Two balls of equal masses are thrown upwards along the same vertical direction at an interval of $2 \mathrm{~s}$, with the same initial velocity of $39.2 \mathrm{~m} / \mathrm{s}$. The two balls will collide at a height of
- A$39.2 \mathrm{~m}$
- Correct answerB$73.5 \mathrm{~m}$
- C$78.4 \mathrm{~m}$
- D$117.6 \mathrm{~m}$
Solution
Let two balls collide at a height $s$ from the ground after $t$ second when second ball is thrown upwards.
$\therefore$ Time taken by first ball to reach the point of collision $=(t+2) \mathrm{s}$
$s=39.2(t+2)+\frac{1}{2}(-9.8)(t+2)^{2}$
$=39.2(\mathrm{t}+2)-4.9(\mathrm{t}+2)^{2}$
For second ball
$\begin{array}{l}
\mathrm{s}=39.2 \mathrm{t}+\frac{1}{2}(-9.8) \mathrm{t}^{2} \\
=39.2 \mathrm{t}-4.9 \mathrm{t}^{2} \quad \text { (ii) } \\
\text { Fromeqs. (i) and (ii) } \\
39.2(\mathrm{t}+2)-4.9(\mathrm{t}+2)^{2}=(39.2) \mathrm{t}-4.9 \mathrm{t}^{2} \\
\text { On solving we get, } \mathrm{t}=3 \mathrm{~s} \\
\text { From Eq. (ii), } \\
\mathrm{s}=39.2 \times 3-4.9 \times(3)^{2}=117.6-44.1=73.5 \\
\mathrm{~m}
\end{array}$
Hence, the correct answer is option: 'B'.