Please explain, question below

Question

A basketball player shoots a basketball with an initial velocity of 15 ft/sec. The ball is released from an initial height of 6.5 feet.

The function         models the height, in feet, of an object after  seconds.  is the initial velocity of the object, and  is the initial height of the object.

Part 1: Write a function that models the height of the basketball. Use your function to answer Parts 2-4.

Part 2: How long does it take for the basketball to hit the ground? Round your answer to the nearest hundredth. Show all of your work.  You're welcome to use this quadratic formula calculator, but please explain your answer. (Links to an external site.)

Part 3: When does the basketball reach its maximum height? Round your answer to the nearest hundredth. Show all of your work and explain your answer.

Part 4: What is the maximum height of the basketball? Round your answer to the nearest hundredth. Show all of your work and explain

Answer 1

Part 1) v₀ denotes the initial velocity of the ball, while h₀ denotes its initial height. So we just need to substitute in the values v₀ = 15 and h₀ = 6.5:

h(t) = -16t² + 15t + 6.5

Part 2) "Hitting the ground" means the height of the basketball goes to zero. So we can use our formula from before, but substituting h(t) = 0.

0 = -16t² + 15t + 6.5

After that, we simply need to solve to t using any method of solving quadratic equations, e.g. that quadratic formula calculator. Using that calculator and plugging in a = -16, b = 15, and c = 6.5, we get the following two solutions:

t = -0.32

t = 1.26

Since it doesn't make sense to talk about a negative time in this context, we throw out the negative solution and we're left with t = 1.26. Therefore, it takes 1.26 seconds for the ball to hit the ground.

Part 3) Since the equation for the height is a quadratic equation with a negative leading coefficient, the shape of the graph is an upside-down parabola. A useful property of parabolas is that the vertex is always exactly halfway between the two zeros. So we just need to find the point halfway between the two zeros that we calculated in Part 2. In other words, we need to find their average:

t_max = (1/2)(1.26 - 0.32) = 0.47

So the basketball is at its maximum height at t = 0.47 seconds.

Part 4) Now we just need to plug the time we found in Part 3 into our original equation for height:

h(t) = -16t² + 15t + 6.5

h(0.47) = -16(0.47)² + 15(0.47) + 6.5

h(0.47) = 10.02

So the maximum height of the basketball is approximately 10.02 ft.

Hope that helps!