Question

A baseball is "popped" straight up by a batter with an initial velocity of 32 ft/sec. The height of the ball above ground is given by a function where t is time in seconds after the ball leaves the bat and h(t) is the height in feet above the ground. The batter hit the ball at an original height of 4 feet off of the ground, and the acceleration due to gravity is -16ft/se2

Answer 1

Answer:

The ball reaches the maximum height after 1 seconds

Step-by-step explanation:

Given

$y = -16t^2 + 32t + 4$ --- the function missing from the question

Required

Time to reach maximum height

The time (t) to reach maximum height is calculated using:

$t = -\frac{b}{2a}$

Where

$y =at^2 + bx + c$

So, by comparison:

$a = -16$

$b = 32$

$c = 4$

So, we have:

$t = -\frac{b}{2a}$

$t = -\frac{32}{2*-16}$

$t = -\frac{32}{-32}$

$t = \frac{32}{32}$

$t = 1$