A tennis ball is thrown from the top of a building. The distance h, in feet, between the tennis ball and the ground, t seconds after it is thrown is given by h(t)=(-16t^2) +160t + 384. At what time is the tennis ball at the maximum height?

Step-by-step explanation:

If the distance h, in feet, between the tennis ball and the ground, t seconds after it is thrown is given by h(t)=(-16t^2) +160t + 384, at maximum height the velocity of the tennis ball will be zero.

Velocity is the rate of change of displacement of a body. If the distance is modeled by the equation h(t)=(-16t^2) +160t + 384 then its velocity will be gotten by taking the derivative with respect to time as shown;

V = dh/dt = -32t+160

at maximum height;

0 = -32t+160

32t = 160

t = 160/32

t = 5seconds

This means that the tennis ball reaches the maximum after 5seconds.

i dont know sorry

step-by-step explanation:

8

step-by-step explanation:

just put √(2) * √(32) into a calculator. that will make the answer be 8.