Question

If a ball is thrown upward at 112 feet per second from the top of a building that is 120 feet high, the
height of the ball can be modeled by s = 120 + 112t - 16t^2where t is the number of seconds after the ball
is thrown. How long after it is thrown is the height 120 feet?

Answer 1

Answer: 7 seconds

Step-by-step explanation:

Obtain the time after the ball is thrown at the height of 120 feet.

Substitute s = 120 in s = -16t^2 + 112t + 120 and solve for t as follows:

$\begin{aligned}&120=-16 t^{2}+112t+120 \\&-16 t^{2}+112 t+120-120=0 \\&-16 t^{2}+112t=0 \\&16 t^{2}-112t=0 \\&16 t(t-7)=0 \\&12 t=0, t-7=0 \\&\Rightarrow t=0, t=7\end{aligned}$

Note that, time cannot be negative or 0. So, ignore t = 0

Therefore, after 7 seconds the ball will reach at a height of 120 feet.