Raylin went to the park with her friends and found a soccer ball. She picked up the soccer ball and kicked it as high as she could. The path of the ball can be represented by h=…

Question

asked Mar 25, 2022 41.6k views

1 Answer

Maysa · Answer 1 · 2022-03-31T12:46:33+0000

Answer:

The maximum height is:
$h(t_(max))=14.39 \: u$

The ball reaches the ground in 1.79 s.

Explanation:

We need to take the derivative and equal to zero to find the time at the maximum height.

h(t)=-16t^(2) +27t+3 (1)

(dh(t))/(dt)=-32t +27=0

t_(max) =(27)/(32)

$t_(max) =0.84\: u$

Now, we just need to put t(max) into equation (1) to find h(max)

h(t_(max))=-16(0.84)^(2) +27(0.84)+3

$h(t_(max))=14.39 \u$

If we want to get the time when the ball reaches the ground we just need to equal h(t) to zero.

0=-16t^(2) +27t+3

Let's solve this quadratic equation.

We will get two solutions and we must choose the positive value.

t1 = 1.79 u

t2 = -0.10 u

Therefore, the ball reaches the ground in 1.79 u.

I hope it helps you!