Question

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

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

Answer 1

Answer: (D)

117.6 m

Answer 2

The scenario is depicted in the diagram below.

A is the starting position for both balls in this figure. B represents the highest point that the initial ball might reach, and C represents the point of impact.
Given in the problem,$u=39.2 \,ms^{-1}$
Time till the 1st ball collides with the 2nd ball (t1)=t sec
Time till the 2nd ball collides with the 1st ball (t2)=(t1−2) sec
Assume the two balls collide t seconds later, with the height of the point of collusion (point) from the ground equal to x.
We obtain by applying the second equation of motion to the first ball
⇒ $s=ut+\frac{at^2}{2}$
⇒ $x=39.2\times t_1+\frac{1}{2}gt_1^{2}$ (In this case, gravity is the only external force acting on the ball.)
⇒ $x= 39.2t_1+\frac{gt_1^2}{2}$ …………. (Equation 1)
We obtain by applying the second equation of motion to the second ball
$S=ut-\frac{1}{2}at^2$
⇒ $x=39.2\times t_2-\frac{1}{2}gt_2^{2}$
⇒ x = $x=39.2 t_2-\frac{1}{2}gt_2^{2}$
⇒$x=39.2\times (t_1-2)-\frac{1}{2}g(t_1-2)^{2}$……….. (Equation 2)
We obtain by substituting the value of x from equation 1 into equation 2.
$39.2 t_1-\frac{1}{2}gt_1^{2}=39.2\times (t_1-2)-\frac{1}{2}g(t_1-2)^{2}$
⇒ $39.2 t_1-\frac{1}{2}gt_1^{2}=$ $39.2t_1-78.4-\frac{1}{2}gt_1^2-2g+2gt_1$
⇒ $2g(t_1-1)=78.4$
⇒ $2\times9.8(t_1-1)=78.4$
⇒ $t_1-1 = 4$
⇒ $t_1 =5 \,sec$
Now, for calculating the distance of the point of collision from the ground (x) , we simply use the equation of speed for the second ball, i.e.
Speed=$\frac{x}{t_2}$
⇒ $v=\frac{x}{t_1-2}$
⇒ $39.2=\frac{x}{3}$
⇒ x=117.6m
Hence, option D is the correct option.