calculus
posted by Brandy .
The height of a projectile launched into the air is given by h (t)= 16t^2 + 48t + 60, where h is the height of the projectile in feet above the ground and t is the number of seconds after its launch. Find the *instantaneous rate of change* in the height of the projectile at 1.5 seconds. It is supposed to be 24 feet per second. Isn't it just 32t + 48 and then putting the 1.5 in for that t?

h(t)=16t^2 + 48t + 60
dh/dt=32t+48, at t=1.5, then
h'=0
you are right. 
yes, but dh/dt=0 at t=1.5
Don't know how they get 24. dh/dt=24 at t=3/4