Question

Two balls of equal masses are thrown upwards along the same vertical direction at an interval of $2s$, with the same initial velocity of $39.2m/s.$ The two balls will collide at a height of

• A. $39.2m$
• B. $73.5m$
• C. $78.4m$
• D. $117.6m$

Answer 1

Answer: (B)

$73.5m$

Given that,

Let the mass of the two body be $m_1=m_2=m$

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 $=(t+2)s$

As per the Newton's second law of motion,

$s=ut+\frac{1}{2}g\times t^2$
$s=39.2(t+2)+\frac{1}{2}(-9.8)(t+2{)}^2$
$=39.2(t+2)-4.9(t+2{)}^2$ ...(i)
For second ball
$s=39.2t+\frac{1}{2}(-9.8)t^2$
$=39.2t-4.9t^2$ ...(ii)
From eqs. (i) and (ii)
$39.2(t+2)-4.9(t+2{)}^2=(39.2)t-4.9t^2$
On solving we get, $t=3s$
From Eq. (ii),
$s=39.2\times 3-4.9\times (3{)}^2=117.6-44.1=73.5m$

so at the height of $73.5m$ these two balls will collide.