Single Choice

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$$39.2m$$
B$$73.5m$$
Correct Answer
C$$78.4m$$
D$$117.6m$$

Solution

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,
$$s=ut+\dfrac{1}{2}g\times{t^2}$$
$$s=39.2(t+2)+\dfrac{1}{2}(-9.8)(t+2)^2$$
$$=39.2(t+2)-4.9(t+2)^2$$ ...(i)
For second ball
$$s=39.2t+\dfrac{1}{2}(-9.8)t^2$$
$$=39.2t-4.9t^2$$ ...(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),
$$s=39.2\times 3-4.9\times (3)^2=117.6-44.1=73.5 m$$
so at the height of $$73.5 m$$ these two balls will collide.

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