1. ## instantaneous velocity

a ball is thrown up in the air. it's height from the ground in meters after t seconds is modeled by h(t)=-5t^2+20t+1. what is the instantaneous velocity of the ball at t=2 seconds?

i calculated h(2)=21 and h(2+h)=-5h^2+21 which i'm not sure are right

what do i do from there?

2. Have you learned derivatives yet? If so, then the derivative of h(t) would be
$\displaystyle h'(t) = -10t + 20$
and you can plug in t = 2.

If not, have you learned limits? Then alternatively you could find
$\displaystyle \lim_{h \to 0} \dfrac{h(2 + h) - h(2)}{h}$
$\displaystyle \lim_{h \to 0} \dfrac{-5h^2 + 21 - 21}{h}$
and take the limit. (Note that the h's are not the same, because I'm trying to be consistent with what you wrote.)

3. i get -5h^2/h=-5(0)^2=0 so it is 0?

4. Close.
$\displaystyle \lim_{h \to 0} \dfrac{-5h^2 + 21 - 21}{h} = \lim_{h \to 0} \dfrac{-5h^2 }{h} = \lim_{h \to 0} -5h = -5(0) = 0$

5. Does this mean that the instantaneous velocity at t=2 seconds is 0 m/s? It doesn't sound very "right to me"

6. Originally Posted by euclid2
Does this mean that the instantaneous velocity at t=2 seconds is 0 m/s? It doesn't sound very "right to me"
Sure it does. When you throw a ball up into the air, the velocity (I'm assuming that upwards is positive) will decrease. The ball slows down traveling upward, and it stops for an instant before traveling downward. So right before the ball falls back down, the velocity is 0 m/s.