Answer
$80feet/sec$
Work Step by Step
We have to calculate initial speed of the ball thrown upwards. As it is represented as a quadratic equation the ball reaches its maximum speed (which is initial speed) $2$ times: at the moment it's thrown and and the moment it falls on the ground.
We have the following equation:
$16t^2-v_{0}t+h=0$
$$Solution 1$$
For a moment, let's forget about the ball going upwards and let's consider when the ball starts falling from $100feet$, where it has initial speed of $0$ and use formula for falling object:
(Note: the maximum speed of the object will be the same)
$h=-16t^2+h_{0}$
$0=-16t^2+100$
$16t^2=100$
$t^2=\sqrt{\frac{100}{16}}=\frac{10}{4}=2.5sec$
Now, input the $t$ value in the equation:
$16\times6.25-v_{0}\times2.5+100=0$
$v_{0}=80ft/s$
$$Solution 2$$
In the equation $16t^2-v_{0}t+100=0$
For a maximum height of $100feet$ (Where height gets only one value) Discriminant should be $0$:
$-{v_{0}}^2-4\times16\times100=0$
${v_{0}}^2=6400$
$v_{0}=80ft/s$
