Question

Exercise Problems.

A ball is gently dropped from a height of 20 m. If its velocity increases uniformly at the rate of 10 ms^–2, with what velocity will it strike the ground? After what time will it strike the ground?

Answer 1

Given: height = 20 m
acceleration = 10 ms^{– 2}
Formula: For free falling body,

1. v = gt
2. s = 1/2 gt²
   Solution:
   time taken to strike the ground
   s = 1/2 gt²
   20m = 1/2 × 10 ms ^{– 2} × t²
   t² = `"40m"/"10ms"^(−2)`= 4s ²
   ∴ t = 2s
3. Velocity of the ball when it strikes ground v = gt
   v = 10ms^{– 2} × 2s
   v = 20ms^{– 1}