A ball is projected vertically upward with an initial velocity of 50 ms⁻¹ at t = 0s. At t = 2s, another ball is projected vertically upward with same velocity. At t = __________ s, second ball will meet the first ball (g = 10 ms⁻²).

Solution

Physics sir · 4 steps

Both balls follow the same path, so they meet where their times sit symmetrically about the top at 5 s.

1. Write both heights. Take upward as positive and measure time t from the first launch. The first ball has been flying for t seconds and the second for (t-2) seconds.

   50t - 5t² = 50(t-2) - 5(t-2)²

2. Solve. The right side expands to 50t - 100 - 5t² + 20t - 20. The 50t and -5t² terms cancel on both sides, leaving 0 = 20t - 120, so t = 6 s.
3. Check. At t = 6 s the first ball is at 300 - 180 = 120 m, falling back after its top at 5 s. The second ball, 4 s into its flight, is at 200 - 80 = 120 m, still rising.
4. So. The second ball meets the first at t = 6 s.

The common slip: 4. Measuring time from the second ball's launch gives 4, not 6. The question's clock started when the first ball was thrown.

The idea to keep. Two balls thrown up with the same speed reach equal heights at times symmetric about the time of the top. With a gap Δ t, they meet at u/g + (Δ t)/2 from the first launch.

Answer from the NTA final answer key. Solution by PrepWiser.