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

## Answer this Question

## Related Questions

- math - A projectile is fired directly upward with a muzzle velocity of 860 feet ...
- College Algebra - A projectile is thrown upward with an initial velocity of 272 ...
- Physics - A projectile is fired up into the air at a speed of 164 m/s at an ...
- Pre-Calculus - Hello, A projectile is launched straight up from the ground with ...
- intermediate algebra - A projectile is launched directly upward from the top of ...
- Intermediate Algebra - If a projectile is fired straight upward from the ground ...
- math - A projectile is launched vertically upward from the top of a 240-foot ...
- pre-calc - analyzing the motion of a projectile: a projectile is fired from a ...
- math - If a projectile is launched from a platform 30 feet high with an initial ...
- physics - three more questions: 1) A 1.50kg projectile is launched at 18.0m/s ...