A ball is thrown upward from 3 feet iff the ground at an initial velocity of 64 feet per second. What is the maximum height the ball reaches?

I found h max = 64 feet

Explanation: Ok...probably you can do this differently but I would try to find the vertex of the parabola describing the trajectory: 1) derive it: h ` ( t ) = 64 − 32 t 2) set derivative equal to zero: 64 − 32 t = 0 t = 64 32 = 2 sec 3) use this value of t into your trajectory: h ( 2 ) = h max 64 ⋅ 2 − 16 ⋅ 4 = 64 feet .

the answer is 45 i just did it on oddesy ware