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.

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

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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 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.

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:

Therefore, the initial velocity would be .

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2(2x + 3) = -4 + 2(2x + 5)

4x + 6 = -4 + 4x + 10

4x + 6 = 4x + 6

4x - 6 + 6 = 4x - 6 + 6

4x = 4x

x = x

unlimited solutions.