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. PLEASE HELP

Answer 1

Answer:

1. $h(t)=-16t^2+15t+6.5$

2. The ball hits the ground at  t = 1.26 seconds

3. The basketball reaches its maximum height at t=0.47 seconds

4. The maximum height of the ball is 3.52 feet

Step-by-step explanation:

Function Modeling

Reality can sometimes be modeled by mathematics. Functions are a great tool to explain the behavior of the measured magnitudes, it can also be used to predict future values and help to make decisions.

We are given a function to model the height (in feet) of a basketball once it's shot from the player. The function is

$h(t)=-16t^2+v_ot+h_o$

where t is the time in seconds, $v_o$ the initial speed and $h_o$ the initial height. We are also given the values

Part 1

$v_o=15\ ft/s,\ h_o=6.5\ ft$

The complete model is

$h(t)=-16t^2+15t+6.5$

Part 2

To find the time the basketball hits the ground, we must set its height to zero:

$-16t^2+15t+6.5=0$

To solve this quadratic equation, we'll use the solver formula

$\displaystyle x=\frac{-b\pm \sqrt{b^2-4ac}}{2a}$

where a = -16, b = 15, c = 6.5

$\displaystyle t=\frac{-15\pm \sqrt{15^2-4\times (-16)\times 6.5}}{2(-16)}$

$\displaystyle t=\frac{-15\pm 25.32}{-32}$

This produces two solutions:

$t=-0.322\ sec,\ t=1.26\ sec$

We discard the negative solution because time cannot be negative, thus the ball hits the ground at  t = 1.26 seconds

Part 3

A quadratic function of the form

$at^2+bt+c$

has its extrema value (maximum or minimum) at

$\displaystyle t=-\frac{b}{2a}$

If a>0, it's a maximum, otherwise it's a minimum

. Since a=-16, we'll get a maximum.

Computing the value of t to make the height be maximum

$\displaystyle t=-\frac{15}{2(-16)}=0.47\ sec$

The basketball reaches its maximum height at t=0.47 seconds

Part 4

The maximum height can be computed by using the function of h evaluated in t=0.47 sec

$h(t)=-16t^2+15t+6.5$

$h_m=-16(0.47)^2+15\times 0.47+6.5$

$h_m=3.52\ feet$

The maximum height of the ball is 3.52 feet