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A ball is gently dropped from a height of $20\ m$. If its velocity increases uniformly at the rate of $10\ ms^{-2}$, 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.