If a ball is thrown directly upward with a velocity of 28 feet per second, its height (in feet) after t seconds is given by y=28t−16t^2. What is the maximum height(feet) attained by the ball?

Question

I need help with this problem! If you could please explain step by step! THank you!

Answer 1

y = s(t) = 28t - 16t²

Please note that this equation tacitly assumes the ball was thrown from an initial height of 0!

The maximum height will occur when the derivative = 0, i.e. 32t=28

Solve this simple equation and plug the answer into the equation for s(t).