Posted by **Brandy** on Friday, April 26, 2013 at 3:49pm.

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?

- calculus -
**bobpursley**, Friday, April 26, 2013 at 3:56pm
h(t)=-16t^2 + 48t + 60

dh/dt=-32t+48, at t=1.5, then

h'=0

you are right.

- calculus -
**Steve**, Friday, April 26, 2013 at 3:58pm
yes, but dh/dt=0 at t=1.5

Don't know how they get 24. dh/dt=24 at t=3/4

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