anonymous
  • anonymous
The height of an object that has been thrown upward from an initial height of 80m at a velocity of 29.4m/s is given by h= -4.9t^2+29.4t+80. At what time is the object at its maximum height?
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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jamiebookeater
  • jamiebookeater
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anonymous
  • anonymous
The object is at it's max height when the derivative of h is equal to 0 (where it is neither rising nor falling) \[dh/dt = -9.8t+29.4\] \[0 = -9.8t+29.4\] Where t = 3, dh/dt = 0. Now you have the time at which the max height occurs you can plug it into the height formula. \[h=-4.9t^2+29.4t+80\] \[h=-4.9*3^2+29.4*3+80\] \[h=124.1\]
anonymous
  • anonymous
and don't forget units ;)
anonymous
  • anonymous
thanks

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