A ball is dropped from the top of a 100 m high tower on a planet. In the last 1/2s before hitting the ground, it covers a distance of 19 m. Acceleration due to gravity (in ms –2 ) near the surface on that planet is _____.

Solution

Physics sir · 5 steps

Call the total fall time T and write two equations: one for the full 100 m, one for the last half second.

1. Full fall. The ball starts from rest, so the whole drop gives 100 = 1/2gT², that is gT² = 200.
2. Last half second. The distance in the last 0.5 s is the total drop minus the drop in the first (T - 0.5) seconds: 1/2gT² - 1/2g(T-0.5)² = gT/2 - g/8. Set this equal to 19 and multiply by 8 to get 4gT - g = 152.
3. Eliminate g. Put g = 200/T² into 4gT - g = 152. This gives 800/T - 200/T² = 152, which tidies into the quadratic below.

   19T² - 100T + 25 = 0 ⇒ T = (100 ± 90)/38 = 5 s or 0.26 s

4. Pick the root. T = 0.26 s is shorter than the half second it must contain, so it is rejected. With T = 5 s, g = 200/25 = 8 m s⁻².
5. So. The acceleration due to gravity on that planet is 8 m s⁻². Check: in the last half second it falls (8 × 5)/2 - 8/8 = 20 - 1 = 19 m.

The common slip: 2. Taking the interval as 1 s instead of half a second gives 2.

The idea to keep. For a distance covered in the last part of a fall, take total time T as the unknown and subtract the drop up to (T - t). Always reject a root where T is shorter than that last interval.

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