The maximum height reached by a projectile is 64 m. If the initial velocity is halved, the new maximum height of the projectile is ______ m.

Solution

Physics sir · 3 steps

Height goes as the square of speed, so halving the speed cuts the height to one quarter.

1. Height formula. At the top, only the vertical velocity has gone to zero. Using v² = u² - 2gH on the vertical motion gives the maximum height.

   H = (u² sin² θ)/2g

2. What changes. The angle θ and g stay the same, so H ∝ u². Halving u multiplies H by (1/2)² = 1/4.
3. So. New height = 64/4 = 16 m.

The common slip: 32. Treating height as proportional to speed and halving 64 gives 32, not 16. Height goes as u², so it drops by a factor of 4.

The idea to keep. For a fixed angle, maximum height, range and time of flight all depend on speed: H and R go as u², T goes as u. Scale by the right power instead of recalculating.

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