A ball is thrown vertically upwards from the ground. It crosses a point the height of $25m$ twice at an interval of $4s$. The ball was thrown at a velocity of:
$A.20m{{s}^{-1}}$
$B.25m{{s}^{-1}}$
$C.30m{{s}^{-1}}$
$D.35m{{s}^{-1}}$

Hint: The equation of time of flight is to be used here.
$T=\dfrac{2u}{g}$
Where $u$ the velocity of the ball is, $g$ is the acceleration due to gravity,$T$ is the time of flight of the ball. And also the height is calculated using the equation,
$H=\left( \dfrac{{{u}^{2}}}{2g} \right)$
Substituting values in this will give us the correct answer for the question.

Complete step by step answer: First of all let us get to know what time of flight and height of the projectile. The time of flight is the time required by the projectile to reach maximum height in a projectile motion.. The maximum height of an object in a projectile trajectory is taken as the vertical component of velocity, it is equal to zero. As the projectile is moving upwards it goes in opposite to gravity, hence the velocity starts to decelerate.
Here in this question, it is given that \[T=4s\]and \[H=25m\]
At a point which is above\[H=25m\], it takes\[T=4s\].
Therefore,
\[4=\dfrac{2u}{g}\]
Hence we will get,
\[u=20m{{s}^{-1}}\]
That means at \[H=25m\] it is having a velocity of\[u=20m{{s}^{-1}}\]. Above this height we can write that,
\[\dfrac{{{u}^{2}}}{2g}=20m\]
Therefore maximum height,
\[20+25=45m\]
With these data, the initial velocity can be calculated as
\[45=\dfrac{{{u}^{2}}}{2g}\]
Hence the initial velocity,\[u\] is
\[u=30m{{s}^{-1}}\]

So the correct answer for this question is option C.

Note: Generally the time of flight is twice the time to reach the maximum height. At maximum height kinetic energy is zero. The time of flight is determined by the initial velocity in the vertical direction and the acceleration due to gravity.