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

The ball reaches the maximum height after 1 seconds

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 = -(b)/(2a)`

Where

`y =at^2 + bx + c`

So, by comparison:

`a = -16`

`b = 32`

`c = 4`

So, we have:

`t = -(b)/(2a)`

`t = -(32)/(2*-16)`

`t = -(32)/(-32)`

`t = (32)/(32)`

`t = 1`