Question

A ball is gently dropped from a height of \(20\) \(m\). If its velocity increases uniformly at the rate of \(10\) \(m s^{-2}\), with what velocity will it strike the ground? After what time will it strike the ground?

Answer 1

Let us assume, the final velocity with which ball will strike the ground be \('v'\) and time it takes to strike the ground be \('t'\)
Initial Velocity of ball, \(u\) =\(0\)
Distance or height of fall, \(s\) = \(20\) \(m\)
Downward acceleration, \(a\) =\(10\) \(m s^{-2}\)
As we know, \(2as =v^2-u^2\)
\(v^2 = 2as+ u^2\)
= \(2 \times 10 \times 20 + 0\)
= \(400\) 
\(\therefore\) Final velocity of ball, \(v\) = \(20\; ms^{-1}\)
\(t = (v-u)/a\)
∴Time taken by the ball to strike = \(\frac{(20-0)}{10}\)

= \(\frac{20}{10}\)

= \(2\) seconds