Question

NEED ANSWERS PLEASE!!
A tee box is 64 feet above its fairway. When a golf ball is hit from the tee box with an initial vertical velocity of 48 ft/s, the quadratic equation 0= -16t^2 + 48t + 64 gives the time, t, in seconds when a golf ball is at height 0 feet on the fairway.
A) Solve the quadratic equation by factoring to see how long the ball is in the air.
B) What is the height of the ball at 1.5 seconds?
C) Is the ball at its maximum height at 1.5 seconds? Explain.

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.