Question

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

height of the ball h in feet after t seconds is given by the equation h= - 16t²-20t+120. How long after the
ball is thrown will it strike the ground?

Answer 1

`t=(-5+√(505))/(8)`

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=(-5 \pm √(505))/(8)`

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

So,

`t=(-5+√(505))/(8)`