From 18 m height above the ground a ball is dropped from rest. The height above the ground at which the magnitude of velocity equal to the magnitude of acceleration (in the same set of units) due to gravity is m. (Take g=10 m/s2 and neglect the air resistance)
Answer
13
Solution
Required speed: v=10 m/s (numerically equal to g=10 m/s2)
Ball dropped from rest: v2=2gs ⇒ 100=2 10 s ⇒ s=5 m (distance fallen)
∴ Height above the ground =18-5=13 m
Answer: 13