April shoots an arrow upward into the air at a speed of 32 feet per second from a platform that is 11 feet high. The height of the arrow is given by the function h(t)equalsminus16t squaredplus32tplus11, where t is the time is seconds. What is the maximum height of the arrow?

27 ft

the maximum height of the arrow is 27 ft

Step-by-step explanation:

Given;

The height of the arrow is given by the function;

h(t) = -16t^2 + 32t + 11

Maximum height is at point when dh(t)/dt = 0.

Differentiating h(t), we have;

dh/dt = -32t + 32 = 0

Solving for t;

-32t = -32

t = -32/-32 = 1

t = 1 (time at maximum height is t = 1)

Substituting t=1 into h(t), to determine the value of maximum height;

h(max)= -16(1^2) + 32(1) + 11

h(max) = 27 ft

the maximum height of the arrow is 27 ft.

hello! just here for points! though i wish i knew!