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An object is thrown straight up with a speed of v(t)=−32t+96 (ft/s)v(t) = -32t + 96 \, (\text{ft/s}), where tt is the time in seconds, from a height of 64 ft64 \, \text{ft} above the ground. What is the maximum height above the ground that the object reaches?

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Q. An object is thrown straight up with a speed of v(t)=−32t+96 (ft/s)v(t) = -32t + 96 \, (\text{ft/s}), where tt is the time in seconds, from a height of 64 ft64 \, \text{ft} above the ground. What is the maximum height above the ground that the object reaches?
  1. Set Velocity to 00: The velocity function is v(t)=−32t+96v(t) = -32t + 96. To find the time when the object reaches the maximum height, we set the velocity to 00.0=−32t+960 = -32t + 96
  2. Solve for t: Solve for t: 32t=9632t = 96\newlinet=9632t = \frac{96}{32}\newlinet=3t = 3 seconds.
  3. Find Height Function: Now we need to find the height at t=3t = 3 seconds. The height function h(t)h(t) is the integral of the velocity function. So, h(t)=−16t2+96t+h(t) = -16t^2 + 96t + initial height.
  4. Calculate Initial Height: The initial height is 64 ft64\,\text{ft}. So, h(t)=−16t2+96t+64h(t) = -16t^2 + 96t + 64.
  5. Plug in t=3t = 3: Plug in t=3t = 3 into h(t)h(t): h(3)=−16(3)2+96(3)+64h(3) = -16(3)^2 + 96(3) + 64.
  6. Calculate h(3)h(3): Calculate h(3)h(3): h(3)=−16(9)+288+64.h(3) = -16(9) + 288 + 64.\newlineh(3)=−144+288+64.h(3) = -144 + 288 + 64.\newlineh(3)=208h(3) = 208 feet.

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