From 4 feet above a swimming pool Susan throws a ball up word with a velocity of 32 ft./s. The height h (t) of the ball t seconds after Susan throws it is given by h (t) = -16t^2+32t+4. For >_0 find the maximum height reached by the ball at the time that the height is reached.

h (t) = 20 ft

Step-by-step explanation:

The equation for the trajectory is according to problem statement

h (t) = - 16*t² + 32*t + 4

Then velocity is dh/dt

d (h) / dt = - 32*t + 32

at h max d (h) / dt = 0

- 32*t + 32 = 0

t = 1 sec

Therefore after 1 second the ball reachs the maximum height

plugging this value in the equation of trajectory we get maximum height

h (t) = - 16*t² + 16*t + 4

h (t) = - 16 * (1) ² + 32*1 + 4

h (t) = 20 ft from swimming pool level