# Question #79341

Feb 13, 2015

${V}_{1} = 26.34 m / s$

${V}_{2} = 29.2 m / s$

I have assumed that by ${V}_{1}$ and ${V}_{2}$ you mean the final velocity of the ball in each case.

We can use this equation of motion:

${v}^{2} = {u}^{2} + 2 a s$

$v =$ final velocity

$u =$ initial velocity

$a =$ acceleration which in this case I will assume to be equal to the acceleration due to gravity which is given by $g = 9.8 m / s / s$.

$s =$ displacement (how far the ball drops).

In the first case:

${V}_{1}^{2} = 0 + \left(2 \times 9.8 \times 35.4\right) = 693.84$

${V}_{1} = \sqrt{693.84} = 26.34 m / s$

In the second case the initial velocity $u = 12.6 m / s$

So:

${V}_{2}^{2} = {\left(12.6\right)}^{2} + \left(2 \times 9.8 \times 35.4\right) = 852.76$

${V}_{2} = \sqrt{852.76}$

${V}_{2} = 29.2 m / s$