Respuesta :
Answer:
[tex]v=\pm48[/tex]
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 [tex]h=-16t^2+vt +4[/tex] 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.
[tex]t=-\frac{b}{2a}=-\frac{v}{2(-16)}=\frac{v}{32}[/tex]
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:
[tex]40=-16(\frac{v}{32} )^2+v\cdot\frac{v}{32} +4[/tex]
[tex]-16\frac{v^2}{1024}+\frac{v^2}{32}+4=40[/tex]
[tex]-\frac{v^2}{64}+\frac{v^2}{32}-36=0[/tex]
[tex]-\frac{v^2}{64}\cdot 64+\frac{v^2}{32}\cdot 64-36\cdot 64=0[/tex]
[tex]-v^2+2v^2-2304=0[/tex]
[tex]v^2-2304+2304=0+2304[/tex]
[tex]v^2=2304[/tex]
[tex]v=\pm\sqrt{2304}[/tex]
[tex]v=\pm48[/tex]
Therefore, the initial velocity would be [tex]\pm48[/tex].
The initial velocity for which the ball reaches a maximum height of 40 ft is; v = ±48 ft/s.
We are given the function that models the height as;
h = -16t² + vt + 4
Now,the formula for the x-coordinate of the parabola which is the point of maximum height is;
t = -b/2a
Thus;
t = -v/(2 × -16)
t = v/32
Put v/32 for t in the height equation;
h = -16(v/32)² + v(v/32) + 4
h = -16v²/1024 + v²/32 + 4
At maximum height, we have;
-16v²/1024 + v²/32 + 4 = 40
-16v²/1024 + v²/32 + 4 - 40 = 0
-16v²/1024 + v²/32 - 36 = 0
v²(1/32 - 16/1024) = 36
0.015625v² = 36
v² = 36/0.015625
v² = 2304
v = √2304
v = ±48 ft/s
Read more about projectile motion at; https://brainly.com/question/10837575
