# Thread: Falling object

1. ## Falling object

A coin is dropped from a height of 750 feet.The height, s, (measured in feet), at time, t (measured in seconds), is given by $s = - 16t^2 +750$

• Find the average velocity on the interval [1,3]
• Find the instantaneous velocity when t = 3
• How long does it take for the coin to hit the ground?
• Find the velocity of the coin when it hits the ground.

2. Originally Posted by Samantha
A coin is dropped from a height of 750 feet.The height, s, (measured in feet), at time, t (measured in seconds), is given by $s = - 16t^2 +750$

You are expected to know that:

$s(t) =$ position function
$v(t) = s'(t) =$ velocity function
$a(t) = v'(t) =$ acceleration function

Given: $s(t) = -16t^2 + 750$

$\Rightarrow v(t) = -32t$

$\Rightarrow a(t) = -32$
• Find the average velocity on the interval [1,3]
The average velocity is given by the slope of the secant line connecting the two points.

Thus, $\mbox { Average Velocity } = \frac {s(3) - s(1)}{3 - 1}$

Can you take it from there?

• Find the instantaneous velocity when t = 3
We simply evaluate v(3), that is, the velocity function when t = 3. I found that for you above

Can you take it from there?

• How long does it take for the coin to hit the ground?
s(t) gives the position function. we want the time when the position is zero, therefore, we set s(t) = 0 and solve for t, this will give you the required time. To be more explicit, we want t such that:

$-16t^2 + 750 = 0$

Can you take it from here?

• Find the velocity of the coin when it hits the ground.
For this question, we simply plug in the value of t we found in the previous question into the velocity function

Can you take it from here?

3. I'm very bad in it...

Can you plz finish the problem.

I would really appreciate...thank you very much

4. How long does it take for the coin to hit the ground?
$s = -16t^2 + 750$

Where s = 0 ; that is where the ground is.

$0 = -16t^2 + 750$

$16t^2 = 750$

Thus $t = +6,85$ or $t = -6,85$

But $t = -6,85$ cannot be true.

Therefore only $t = +6,85$ is true.

5. Find the velocity of the coin when it hits the ground.
We know it hits the ground in 46,875 seconds. We calculated that above.

But now we need to differentiate $s$

Remember the following:
When S(distance) is differentiated, we get V(velocity).
When V is differentiated, we get A(acceleration).

----------

So:

$\frac{dS}{dT} = V = -32t$

We know t = 6,85 seconds.

So $V = -32(6,85)$

$V = -219,2 \ feet \ per \ second$

What does the negative mean? It merely means the coin is moving downwards.