The equation you need is
s(t) = s(0) + v0t - (1/2)gt2 where g is the acceleration due to gravity.
v(t)=s'(t)
The maximum height occurs when v(t)=0

Paul M.
07/08/22
John R.
asked 07/07/22A ball is thrown upward with an initial velocity of 48 feet per second from an initial height of 64 feet. The acceleration due to gravity is a(t) = −32 feet per second squared.
(a) Find the velocity function, v(t), of the ball at time t.
(b) Find the position function, s(t), of the height of the ball at time t.
(c) At what time does the ball reach its maximum height?
(d) What is the maximum height reached by the ball?
The equation you need is
s(t) = s(0) + v0t - (1/2)gt2 where g is the acceleration due to gravity.
v(t)=s'(t)
The maximum height occurs when v(t)=0
Paul M.
07/08/22
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John R.
I'm quite confused about that first equation07/07/22