Quadratic formula problem. Please help!

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. (I got Part 1: h(t)= -16+15t+6.5 but i dont know how to do the rest)

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 your answer.

Answer 1

Since you have a good start, I'm going to give you some hints, which I think will get you on your way.

The formula you printed above starts with -16 instead of -16t². I assume this is a typo. If not, you need to fix it. Think about what "hitting the ground" means in this context. What is h in that scenario? Once you see that h = 0 is ground level, you then know that actually solving the quadratic for its zeroes, will answer the question. There should be both a negative and a positive zero. The negative zero is superfluous, since time can't be negative in this context.

The time of achieving maximum height and the height itself can also be pictured. Think of the ball following the path of a negative parabola. Its maximum height is the y of the vertex, and the time it reaches that height is the x of the vertex. As you know, you can find the vertex by completing the square and converting the formula to vertex form. An alternate approach that I like is to use the mini formula for the x of the vertex: -b/2a . This would give us 15/32 or 0.469 seconds. You can then plug that into the formula to get you y value for maximum height.

Best wishes!

Answer 2

Pretty close on the first part, the first term should be -16t².

For the second part, the ball will hit the ground when the height is zero, so set h(t) = 0 and use the quadratic formula to solve for t.

For Part 3, the ball reaches its max height at the vertex of the quadratic. The vertex of a quadratic in the form at² + bt + c is found at t = -b/(2a).

For Part 4, just calculate h(t) at the value you found for t in Part 3.