Hi Kathy,

There is not enough information to determine the initial velocity. The initial velocity depends on the angle.

The parametric equations given in the earlier question are

\[x(t) = (v_0 \cos \theta)t \mbox{ and } y(t) = h + (v_0 \sin\theta)t-16t^2.\]

If $\theta$ is 45 degrees then using the data given and letting $t_1$ be the time when the ball is caught these become

\[x(t) = \left(v_0 \times \frac{1}{\sqrt{2}}\right)t_1 = 90 \mbox{ and } y(t) = 7 + \left(v_0 \times \large \frac{1}{\sqrt{2}} \right)t_1 -16t_1^2 = 4 .\]

As I indicated in my earlier response you can substitute $(v_0 \times \frac{1}{\sqrt{2}})t_1 = 90$ into the second equation and solve for $t_1$ then substitute the value of $t_1$ into $(v_0 \times \frac{1}{\sqrt{2}})t_1 = 90$ and solve for $v_0.$ I got $t_1$ to be a little less than 2.5 seconds and $v_0$ about 53 feet per second. (You should check my arithmetic.)

What if $\theta$ is 30 degrees? This would give

\[x(t) = \left(v_0 \times \frac{\sqrt{3}}{2} \right)t_1 = 90 \mbox{ and } y(t) = 7 +\left (v_0 \times \large \frac{1}{2} \right)t_1 -16t_1^2 = 4 .\]

This time substituting the first equation into the second I got $16 t_1^2 = \normalsize \frac{90}{\sqrt{3}} + 3$ which gave me $t_1$ approximately 1.9 seconds and $v_0$ almost 107 feet per second.

Throwing at a lower arc means the ball gets to the receiver more quickly but you need to throw it much harder.

I hope this helps,
Harley