How high is a baseball after 5 seconds of being thrown off of a 175 foot building? And how many seconds does it take to hit the ground?

Question

A person standing close to the edge on the top of a 175-foot building throws a baseball vertically upward. The quadratic equation

$$h = -16 t^2 + 160 t + 175$$

models the ball's height, $h$ , above the ground in feet, $t$ seconds after it was thrown.

1. How high is the baseball after five seconds?

2. How many seconds does it take until the ball finally hits the ground? Round to the nearest tenth of a second.

Answer 1

HINTS:

(1) Evaluate $$h(5) = -16(5)^2 + 160(5) + 175$$ I.e. Plug into the formula the number $5$ and calculate $h(5)$: the height of the ball five seconds after it was thrown.

(2) Solve for $t$ when $h(t) = 0$. That is, find the zeros of $$h = -16t^2 + 160 t + 175 = 16t^2 - 160 t - 175 = 0$$ Try to factor to find the zeros of the equation, or else use the quadratic formula to find the zeros: what values of $t$ make $h$ equal zero? Which value is positive? That's the time in seconds from when the ball was thrown until when the ball hits the ground. You may hover over the solution below to check your answer:

$$t = 5 + \dfrac{5\sqrt {23}}{4} \approx \;11\;\;\text{seconds}$$