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

The Correct Option is D

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.