Question

A ball is thrown downward from the top of a 120-foot building with an initial velocity of 20 feet per second. The

Answer 1

Answer:

$t=\frac{-5+\sqrt{505}}{8}$

Step-by-step explanation:

The ball strikes the ground when h = 0.

$-16t^2 - 20t + 120 = 0 \\ \\ 4t^2 + 5t - 30=0 \\ \\ t=\frac{-5 \pm \sqrt{5^{2}-4(4)(-30)}}{4(2)} \\ \\ t=\frac{-5 \pm \sqrt{505}}{8}$

However, as time most be positive, we only consider the positive case.

So,

$t=\frac{-5+\sqrt{505}}{8}$