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# 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?

Given:

Initial Velocity of the ball, $u=0$

Distance or height of fall, $s=20m$

Downward acceleration, $a=10\ ms^{-2}$

To find: Final velocity with which the ball will strike the ground, $(v)$, and time taken by the ball to strike the ground, $(t)$

Solution:

Suppose, the final velocity with which the ball will strike the ground be $‘v’$ and the time taken by it to strike the ground be $‘t’$.

By the third equation of motion, we know that-

$v^2=u^2+2as$

Putting the given values, we get-

$v^2=0^2+2\times {10}\times {20}$

$v^2=0+400$

$v^2=400$

$\sqrt {v^2}=\sqrt {400}$

$v=20m/s$

Thus, the final velocity with which the ball will strike the ground is 20m/s.

Now,

By first equation of motion, we know that-

$v=u+at$

Putting the required values, we get-

$20=0+10\times t$

$20=10t$

$t=\frac {20}{10}$

$t=2s$

Thus, time taken by the ball to strike the ground is, 2 seconds.

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