A ball is thrown into the air with an upward velocity of 38 ft/s. Its height h(t) in feet after t seconds is given by the function h(t)=-16^2+38t+7.
A. What is the initial heights?
B. Find the height of the ball at 1.5 seconds
C. After how many seconds will the ball reach its maximum height?
D. What is the ball's maximum height?
E. When will the ball hit the ground? (Round to the tenth place) Show or explain how you got your answer.

In a quadratic modeling free fall, h(t) = -16t² + v₀t + h, h(t) is the height of the object AFTER the fall while h is the initial height of the object. To answer a:

a. The initial height of the object is 7 feet

To find the height, h(t), after 1.5 seconds for b., evaluate the quadratic at h(1.5):

h(1.5) = -16(1.5)² + 38(1.5) + 7 so

b. h(1.5) = 29.5 feet

In order to determine the max height of the object, we need to put this quadratic into vertex form. This is an upside down (negative) parabola, so the vertex represents a max height. It is at the vertex values of (h, k) that we will find the answers to both c and d. To put this into vertex form, we need to complete the square on the quadratic. Do this by first setting the quadratic equal to 0 then moving over the constant to get:

-16t² + 38t = -7

The first rule for completing the square is that leading coefficient has to be a positive 1. Ours is a -16, so we have to factor it out:

-16(t² - 2.375t) = -7

The next rule for completing the square is to take half the linear term, square it, and then add that to both sides. Our linear term is 2.375. Half of that is 1.1875, and 1.1875 squared is 1.41015625. BUT we cannot forget about that -16 sitting out front of those parenthesis. It is a multiplier. That means that we did not just add in 1.41015625, we added in -16(1.41015625):

-16(t² - 2.375t + 1.41015625) = -7 - 22.5625

The reason we do this is to create a perfect square binomial on the left. Writing the left side in terms of our binomial and at the same time simplifying the right:

-16(t - 1.1875)² = -29.5625

We finish it off by adding 29.5625 to both sides and setting the vertex form of the quadratic back equal to y:

y = -16(t - 1.1875)² + 29.5625

From this we can ascertain that the vertex is located at (1.1875, 29.5625)

The answer to c is found in the h coordinate of the vertex which 1.1875.

c. 1.1875 seconds to reach its max height

The answer to d is found in the k coordinate of the vertex which is 29.5625

d. The max height is 29.5625 feet

For e, we will look back to the original function. We determined that h(t) is the height of the object AFTER it falls. If the object falls to the ground, it goes to follow that the height of the object while it is on the ground is 0. We sub in 0 for h(t) and factor to find the times at which the object hits the ground.

Plugging our values into the quadratic formula gives us the times of -.017 seconds and 2.5 seconds. Since we know that time can't EVER be negative, the object hits the ground 2.5 seconds after it was dropped.

Good luck with your quadratics!