Question

A six-foot-tall basketball player shoots a ball towards the basket and misses. The height in feet of the ball up off the floor of the court as it travels through the air can be modeled by the function: h(z) =-16x2 + 48z + 6. Where x is the number of seconds the ball is in the air after the shot is taken. When does the ball reach its maximum height? What is themaximum height of the ball? How much time will pass in seconds from the start of the shot to the ball hitting the floor? Answer all three questions to receive full credit.

Answer 1

Step 1:

Write the function

`\text{h(x) = -16x}^2\text{ + 48x + 6}`

Step 2:

When does the ball reach its maximum height? mean you should calculate the time taken for the ball to reach maximum height.

The ball reach maximum height at v = 0, when velocity = 0 m/s

Step 3:

`\begin{gathered} h(x)=-16x^2\text{ + 48x + 6} \\ v\text{ = -32x + 48} \\ -32x\text{ + 48 = 0} \\ 32x\text{ = 48} \\ x\text{ = }(48)/(32) \\ \text{ x = }(6)/(4) \\ \text{x = }(3)/(2)\text{ = 1.5 seconds} \end{gathered}`

a) the ball reaches its maximum height at x = 1.5 seconds

b)

maximum height reach at x = 1.5 seconds

`\begin{gathered} m\text{aximum height} \\ =-16(1.5)^2\text{ + 48(1.5) + 6} \\ =\text{ -36 + 72 + 6} \\ =\text{ 42} \\ \end{gathered}`

The maximum height of the ball is 42m

c)

the time that will pass in seconds from the start of the shot to the ball hitting the floor = 2 x 1.5

= 3 seconds