Aball is thrown up into the air with an initial velocity of 64 ft/sec. the ball is 4 ft above the ground when it is released. its height, in feet, at time t seconds is given by: h(t) = −16t^2 + 64t + 4. what is the time when it reaches its maximum height?

2 seconds and maximum height reached is 68 ft

Explanation:

h (t) = -16t^2 +64t +4

differentiate with respect to t

h'(t) = -32t +64

for maximum height velocity which is h'(t) must be zero

-32t +64 = 0

-32t = -64

t = 2 s

hence height is

h(2) = -16(2)^2 +64(2) +4

h(2) = -64+128+4

h(2) = 68 feet

i think it’s a. this is because if they measured the mass of something with a mass of 150g, then their final answer would have to be approximately 150g. though all of their answers are almost 150, rosa and tran have the closest answers (or the exact answer) to the accurate one.

0.55x745.7j and a calculator