Question

Suppose a ball is thrown up straight into the air, and the height of the ball above the ground is given by the function h(t)=6+37t-16t^2, where h is in feet and t is in seconds.

a) What is the velocity of the ball at time t=3.2
b) At what time t does the ball stop going up and start returning to earth?

Answer 1

a)
Deriving the function gives us the velocity at any time t: 
h(t) = 6 + 37t - 16t²
h'(t) =  v(t) = 37 - 32t

Plugging in 3.2 for t...
v(3.2) = 37 - 32(3.2) 
v(3.2) = -65.4

b) The point at which the ball stops going up and begins to come down is the point at which its velocity is zero, suspended for a moment in the air before falling. With this in mind, we set v(t) equal to zero and solve for t
v(t) = 37 - 32t
0 = 37 - 32t
32t = 37 
t = 1.16