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An object is thrown straight up with a speed of v(t)=−32t+96v(t) = -32t + 96 ft/s, where tt is the time in seconds, from a height of 6464 ft above the ground. What is the maximum height above the ground that the object reaches?

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Full solution

Q. An object is thrown straight up with a speed of v(t)=−32t+96v(t) = -32t + 96 ft/s, where tt is the time in seconds, from a height of 6464 ft above the ground. What is the maximum height above the ground that the object reaches?
  1. Write Velocity Function: First, let's write down the velocity function: v(t)=−32t+96v(t) = -32t + 96.
  2. Find Time of Zero Velocity: To find the maximum height, we need to find the time when the velocity is zero because that's when the object stops going up and starts falling down. So, set v(t)v(t) to 00 and solve for tt.0=−32t+960 = -32t + 96.
  3. Calculate Time: Solve for tt: 32t=9632t = 96, so t=9632t = \frac{96}{32}.t=3t = 3 seconds.
  4. Determine Position Function: Now we need to find the position function, s(t)s(t), which gives the height of the object at any time tt. We know the initial velocity is 9696 ft/s and the initial height is 6464 ft. The position function is the integral of the velocity function.\newlines(t)=−16t2+96t+64s(t) = -16t^2 + 96t + 64.
  5. Find Maximum Height: Plug in the time we found t=3t = 3 into the position function to find the maximum height.s(3)=−16(3)2+96(3)+64.s(3) = -16(3)^2 + 96(3) + 64.
  6. Calculate Maximum Height: Calculate the maximum height: s(3)=−16(9)+288+64.s(3) = -16(9) + 288 + 64.\newlines(3)=−144+288+64.s(3) = -144 + 288 + 64.
  7. Final Result: Finish the calculation: s(3)=144+64s(3) = 144 + 64.\newlines(3)=208s(3) = 208 feet.

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