Question

WILL MEDAL
A catapult launches a boulder with an upward velocity of 120 ft/s. The height of the boulder h in feet after t seconds is given by the function h=-16t^2+120t+10. What is the boulders maximum height? How long does it take it to reach the maximum height? Round to the nearest hundreth if neccesary.

Answer 1

In here we must take the derivative of the function h. Let me:
\[h'(t) = -16 \times 2 \times t^{2-1} + 120 \times t^{1-1} + 0\]
After this you we find the 't' value which makes this function equal to 0. To do this, we simply make this equation equal to zero:

\[-32t + 120 = 0\]
\[t = 3.75\]

In geometrically what we have done is finding a parabola's highest point. Anyway, put this 't' value in the original equation, we will get the maximum height.

\[h(3.75) = -16 \times (3.75)^2 + 120 \times 3.75 + 10 = 235\]

This is the maximum height. And time elapsed to take this height is 3.75 seconds. Look at some differentiation subjects for more information about these kind of problems.

Answer 2

Thanks!

Answer 3

Here is some material: http://www.statistica.com.au/differentiation_max_and_min.html