Question

Two balls of equal masses are thrown upwards, along the same vertical direction at an interval of 2 seconds, with the same initial velocity of 40 m/s. Then they collide at a height of (Take g=10m/s²)

• A. 120 m
• B. 75 m
• C. 200 m
• D. 45 m

Answer 1

Answer: (B)

75 m

Answer 2

Let $T$ be the time elapsed since the first ball was thrown. The second ball is thrown 2 seconds later, so its time of flight is $T-2$. For the first ball: $h = 40T - 5T^2$ For the second ball: $h = 40(T-2) - 5(T-2)^2$ Equating heights: $40T - 5T^2 = 40(T-2) - 5(T-2)^2$ Solving for $T$: $T = 5$ s. Substituting $T=5$ s into the first ball's height equation: $h = 40(5) - 5(5)^2 = 200 - 125 = 75$ m.