1. Velocity of a Ball

A small rubber ball is thrown into the air by a child at a velocity of 20 feet per second. the child releases the ball 4 feet above the ground. What is the velocity of the ball after 1 second?
modeled by: s(t)=-16t^2+v0t +s0 where v0 is the velocity and s0 is the vertical position of object at time t=0....HUH?

I need help on how to set up. TY!

2. Originally Posted by Snarfsnarf2006
A small rubber ball is thrown into the air by a child at a velocity of 20 feet per second. the child releases the ball 4 feet above the ground. What is the velocity of the ball after 1 second?
modeled by: s(t)=-16t^2+v0t +s0 where v0 is the velocity and s0 is the vertical position of object at time t=0....HUH?

I need help on how to set up. TY!
I assume you are in a Calculus class. The velocity is the first time derivative of the position function.

-Dan

3. further info needed on velocity of a ball

I'm sorry, but I don't understand your statement. Could you kindly explain? Sadly, yes, I'm in calculus...it makes no sense to me.

4. Originally Posted by Snarfsnarf2006
I'm sorry, but I don't understand your statement. Could you kindly explain? Sadly, yes, I'm in calculus...it makes no sense to me.

$v(t) = \frac{ds}{dt}$ is your velocity function. Take the derivative of s(t) to get v(t) and plug in t = 1.

Or are you saying you don't know how to do the derivative?

-Dan

5. in response to

both actually. My brain just doesn't seem to grasp math. Particularly w/letters and #'s. Sorry to be a pest.

6. Originally Posted by Snarfsnarf2006
both actually. My brain just doesn't seem to grasp math. Particularly w/letters and #'s. Sorry to be a pest.
$s_0 = 4$ and $v_0 = 20$, so

$s(t) = -16t^2 + 20t + 4$

$v(t) = -32t + 20$

Thus at t = 1:
$v(1) = -32 \cdot 1 + 20 = -12$

So the velocity is 12 ft/s downward.

Please tell me/us all what you don't understand in the above, if anything.

-Dan