Determine the maximum height of the projectile given by h(t) = -16t^2 + 64t + 80. (2023)

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Q: Determine the maximum height of the projectile given by h(t) = -16t^2 + 64t + 80. (2023)

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

Steps: 12

Step 1: Identify the function for height, which is h(t) = -16t^2 + 64t + 80.
Step 2: Recognize that this is a quadratic function in the form h(t) = at^2 + bt + c, where a = -16, b = 64, and c = 80.
Step 3: To find the time t when the maximum height occurs, use the formula t = -b/(2a).
Step 4: Substitute the values of a and b into the formula: t = -64/(2 * -16).
Step 5: Calculate the denominator: 2 * -16 = -32.
Step 6: Now calculate t: t = -64 / -32 = 2.
Step 7: Now that we have t = 2, substitute this value back into the height function to find the maximum height: h(2) = -16(2^2) + 64(2) + 80.
Step 8: Calculate 2^2, which is 4, then multiply: -16 * 4 = -64.
Step 9: Calculate 64 * 2, which is 128.
Step 10: Now add these results together: h(2) = -64 + 128 + 80.
Step 11: First, add -64 and 128 to get 64, then add 80: 64 + 80 = 144.
Step 12: Therefore, the maximum height of the projectile is 144.