Question

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?

Answer 1

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.
\[h=-4.9t^2+29.4t+80\]
\[h=-4.9*3^2+29.4*3+80\]
\[h=124.1\]

Answer 2

and don't forget units ;)

Answer 3

thanks