Question

a ball is thrown upward with an initial velocity of 48 ft/sec from a height 640 ft. its height h, in feet, after t seconds is given by h(t)= - 16t² + 48t + 640. after how long will the ball reach the ground? the ball will reach the ground in seconds.

Explanation:

Step1: Set height function to zero

The ball reaches the ground when $h(t)=0$. So we set $- 16t^{2}+48t + 640=0$. Divide through by -16 to simplify: $t^{2}-3t - 40=0$.

Step2: Factor the quadratic equation

We factor $t^{2}-3t - 40$ as $(t - 8)(t+5)=0$.

Step3: Solve for t

Using the zero - product property, if $(t - 8)(t + 5)=0$, then $t-8=0$ or $t + 5=0$. So $t=8$ or $t=-5$. Since time cannot be negative in this context, we discard $t=-5$.

Answer:

8