What is the maximum height of the projectile modeled by h(t) = -16t^2 + 64t + 48

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Question: What is the maximum height of the projectile modeled by h(t) = -16t^2 + 64t + 48? (2021)

Options:

Correct Answer: 64

Exam Year: 2021

Solution:

The maximum height occurs at t = -b/(2a) = -64/(2*-16) = 2. h(2) = -16(2^2) + 64(2) + 48 = 80.

What is the maximum height of the projectile modeled by h(t) = -16t^2 + 64t + 48? (2021)

What is the maximum height of the projectile modeled by h(t) = -16t^2 + 64t + 48? (2021)

Step 1: Identify the equation of the projectile's height, which is h(t) = -16t^2 + 64t + 48.
Step 2: Recognize that this is a quadratic equation in the form h(t) = at^2 + bt + c, where a = -16, b = 64, and c = 48.
Step 3: To find the maximum height, use the formula for the time at which the maximum occurs: t = -b/(2a).
Step 4: Substitute the values of b and a into the formula: t = -64/(2 * -16).
Step 5: Calculate the denominator: 2 * -16 = -32, so t = -64 / -32 = 2.
Step 6: Now that we have t = 2, substitute this value back into the original height equation to find the maximum height: h(2) = -16(2^2) + 64(2) + 48.
Step 7: Calculate 2^2, which is 4, then multiply: -16 * 4 = -64.
Step 8: Calculate 64 * 2, which is 128.
Step 9: Now add the results: h(2) = -64 + 128 + 48.
Step 10: Combine the numbers: -64 + 128 = 64, then 64 + 48 = 112.
Step 11: Therefore, the maximum height of the projectile is 112.

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