Question

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.

Answer 1

Answer:

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