Question

suppose a ball is thrown straight up 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. What is the velocity of the ball at time t = 3.2

Answer 1

t = 3.2 seconds I presume
h(t) = 6 + 37t - 16t^2
velocity = dh / dt
v = 0 + 37 - 32 t which is initial velocity - g t where g = 32 ft/s^2 (old text:)
at t = 3.2
v = 37 - 32 (3.2) = 37 - 102.4 = - 70.4 ft/second

by the way at t = 3.2
h = 6 + 37(3.2) - 16(3.2)^2 = 6 + 118.4 - 163.8 = -39.4 feet
below ground :)

Answer 2

-65.4 ft/sec
nDeriv(6 + 37x - 16x^2, x, 3.2)

this gives you the derivative of the height function with respect to time (the velocity) at the time t = 3.2