# Question #01e2a

Mar 1, 2018

We can use $v = u + a t$ to calculate speed of the ball after $t s$,

here, $u = 4 m {s}^{-} 1 , a = g , t = 2$

So, $v = 4 + 2 \cdot 9.8 = 23.6 m {s}^{-} 1$

And,if its speed is $V$ while touching the ground,then we can use, ${v}^{2} = {u}^{2} + 2 g h$

here, $h = 16 m$

So, $V = \sqrt{16 + 2 \cdot 16 \cdot 9.8} = 18.157 m {s}^{-} 1$

Oops! The velocity value while touching the ground is less than the velocity found at $2 s$,that means the ball has reached the ground before that.

Let's calculate time required to reach the ground,using $V = u + g t$

So, $t = \frac{18.157 - 4}{9.8} = 1.45 s$

So,now if we consider that after the impact the ball went upwards,and also consider that coefficient of restitution was $1$,then that ball will go up with the same velocity.

So,for $\left(2 - 1.45\right) = 0.55 s$ while going up,if it achieves a velocity of $v$,then we cans say,

$v = u - g t$

here, $u = 18.157 m {s}^{-} 1$

So, $v = 12.767 m {s}^{-} 1$ (upwards)(this is the velocity at the end of $2 s$)