Question

You hit a ball from 4 ft above the ground with an initial vertical velocity of v ft/s. The function h = −16t2 + vt + 4 models the height h in feet of the ball at time t in seconds. Explain how to find the initial velocity for which the ball reaches a maximum height of 40 ft.

Answer 1

Answer:

$v=\pm48$

Step-by-step explanation:

We have been given that you hit a ball from 4 ft above the ground with an initial vertical velocity of v ft/s. The function $h=-16t^2+vt +4$ models the height h in feet of the ball at time t in seconds.

First of all, we will find the time, when height will be maximum. We know that vertex is the maximum point for a downward opening parabola.

$t=-\frac{b}{2a}=-\frac{v}{2(-16)}=\frac{v}{32}$

To find the initial velocity for which the ball reaches a maximum height of 40 ft, we will substitute height (h) equal to 40 and t is equal to v/32 to  solve for v as:

$40=-16(\frac{v}{32} )^2+v\cdot\frac{v}{32} +4$

$-16\frac{v^2}{1024}+\frac{v^2}{32}+4=40$

$-\frac{v^2}{64}+\frac{v^2}{32}-36=0$

$-\frac{v^2}{64}\cdot 64+\frac{v^2}{32}\cdot 64-36\cdot 64=0$

$-v^2+2v^2-2304=0$

$v^2-2304+2304=0+2304$

$v^2=2304$

$v=\pm\sqrt{2304}$

$v=\pm48$

Therefore, the initial velocity would be $\pm48$.