math
posted by lucy .
If a ball is thrown directly upward with a velocity of 80ft/s, its height (in feet) after t seconds is given by y=80t16t^2. What is the maximum height attained by the ball?

That would be when the velocity is zero. That happens when the derivative
80  32t, is zero. Therefore t = 2.5 seconds at that time, and the maximum height is
80*2.5  16(2.5)^2 = 100 ft 
what ever

Alternatively:
If a ball is thrown directly upward with a velocity of 80ft/s, its height (in feet) after t seconds is given by y=80t16t^2. What is the maximum height attained by the ball?
The time to reach the maximum height and zero velocity derives from Vf = Vo  gt wherer Vf = the final velocity = 0, Vo = the initial velocity = 80fps, t = the time to zero velocity and g = the deleration due to gravity.
Therefore, 0 = 80  32t making t = 2.5 sec.
Then, from h = 80t16t^2, h = 80(2.5)  16(2.5)^2 = 200  100 = 100 ft. 
thanks!