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A baseball is "popped" straight up by a batter with an initial velocity of 32 ft/sec. The height of the ball above ground is given by a function where t is time in seconds after the ball leaves the bat and h(t) is the height in feet above the ground. The batter hit the ball at an original height of 4 feet off of the ground, and the acceleration due to gravity is -16ft/se2

Sagot :

MrRoyal

Answer:

The ball reaches the maximum height after 1 seconds

Step-by-step explanation:

Given

[tex]y = -16t^2 + 32t + 4[/tex] --- the function missing from the question

Required

Time to reach maximum height

The time (t) to reach maximum height is calculated using:

[tex]t = -\frac{b}{2a}[/tex]

Where

[tex]y =at^2 + bx + c[/tex]

So, by comparison:

[tex]a = -16[/tex]

[tex]b = 32[/tex]

[tex]c = 4[/tex]

So, we have:

[tex]t = -\frac{b}{2a}[/tex]

[tex]t = -\frac{32}{2*-16}[/tex]

[tex]t = -\frac{32}{-32}[/tex]

[tex]t = \frac{32}{32}[/tex]

[tex]t = 1[/tex]

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