An object is thrown straight up with a speed of v(t)=−32t+96, where t is the time in seconds, from a height of 64ft above the ground. What is the maximum height above the ground that the object reaches?
Q. An object is thrown straight up with a speed of v(t)=−32t+96, where t is the time in seconds, from a height of 64ft above the ground. What is the maximum height above the ground that the object reaches?
Identify velocity function: Identify the velocity function and initial height. v(t)=−32t+96, initial height h0=64 ft.
Find height function: The height function h(t) is the integral of the velocity function.h(t)=∫v(t)dt=∫(−32t+96)dt.
Integrate velocity function: Integrate the velocity function to find the height function. h(t)=−16t2+96t+C.
Determine constant of integration: Determine the constant of integration C using the initial height. h(0)=64ft, so C=64.
Write complete height function: Write the complete height function. h(t)=−16t2+96t+64.
Find time of maximum height: Find the time at which the object reaches maximum height by finding the vertex of the parabola.The vertex occurs at t=−2ab, where a=−16 and b=96.
Calculate time of maximum height: Calculate the time at which the object reaches maximum height. t=−2(−16)96, t=3296, t=3 seconds.
Substitute time into height function: Substitute t=3 into the height function to find the maximum height.h(3)=−16(3)2+96(3)+64.
Perform calculation for maximum height: Perform the calculation to find the maximum height. h(3)=−16(9)+288+64, h(3)=−144+288+64.
Add values for maximum height: Add the values to get the maximum height.h(3)=144+64, h(3)=208 feet.
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