Question

a ball is thrown and its height is modeled by h = -16t² + 32t + 48. at what time does it hit the ground? t = 1 t = 3 t = 2 t = 4

Explanation:

Step1: Set height h to 0

When the ball hits the ground, its height is 0. So we set $-16t^{2}+32t + 48=0$.

Step2: Divide the equation by -16

Dividing each term by -16 gives $t^{2}-2t - 3=0$.

Step3: Factor the quadratic equation

We factor $t^{2}-2t - 3$ as $(t - 3)(t+ 1)=0$.

Step4: Solve for t

Using the zero - product property, if $(t - 3)(t + 1)=0$, then $t-3=0$ or $t + 1=0$. So $t=3$ or $t=-1$. Since time cannot be negative in this context, we discard $t=-1$.

Answer:

B. $t = 3$