Question

NEED ANSWERS PLEASE!!

Answer 1

(A)  4 sec the ball is in the air.

(B) Height of the ball = 49 ft.

(C) Yes, the ball is at its maximum height at 1.5 seconds.

Solution:

Given data:

$h(t)=-16t^2+48t+64$

Initial velocity = 48 ft/s

Height = 64 ft

(A)  $-16t^2+48t+64=0$

a = –16, b = 48, c = 64

We can solve it by using a quadratic formula,

$\frac{-b \pm \sqrt{b^{2}-4 a c}}{2 a}$

$\Rightarrow t=\frac{-48 \pm \sqrt{(48)^{2}-4 \times(-16)(64)}}{2(-16)}$

$\Rightarrow t=\frac{-48 \pm \sqrt{2304+4096}}{-32}$

$\Rightarrow t=\frac{-48 \pm \sqrt{6400}}{-32}$

$\Rightarrow t=\frac{-48 \pm 80}{-32}$

$\Rightarrow t=\frac{-48 + 80}{-32},\frac{-48 - 80}{-32}$

$\Rightarrow t=-1,t=4$

Time cannot be in negative. So neglect t = –1

t = 4 sec

Hence, 4 sec the ball is in the air.

(B) When t = 1.5 sec,

$h(1.5)=-16(1.5)^2+48(1.5)+64$

h(1.5) = 49 ft

(C) The maximum height occurs at the average of zeros.

Average = $\frac{(-1+4)}{2}=1.5$ sec

Yes, the ball is at its maximum height at 1.5 seconds.