# Math Help - Velocity Question

1. ## Velocity Question

A ball is thrown straight down from the top of a 245 ft building with an initial velocity of -20 ft per second. What is its velocity after 4 seconds? What is its velocity after falling 100 ft?

I have the first part = -148 ft/sec

I cannot figure out what to do for the second part.

2. Derivative of position-time gives velocity-time. Derivative of velocity-time gives acceleration.

Well, we know the acceleration.
$a = 9.8$

So now if we take the integral of acceleration, we get velocity.
$v = 9.8x + c$

And we know that at 0 seconds, we're traveling -20ft/s
$-20 = 9.8(0) + c$
$c = -20$

And so we get:
$v = 9.8x - 20$

The integral of this will then give us distance:
$d = \frac {9.8x^2}{2} - 20x + c$

And we know that we are 245 feet up at 0 seconds.

$245 = \frac {9.8(0)^2}{2} - 20(0) + c$
$c = 245$

And thus our final equation:
$d = \frac {9.8x^2}{2} - 20x + 245$

You may recognize this last equation from physics class.

d = (vi)t + 1/2at^2

In the form of d - 245

3. I understand what you are saying, but I don't understand how to apply the 100 ft...I need to find the velocity...

4. Strange that they'd say a ball was thrown straight down and yet say it started at -20m/s rather than 20m/s. Still proper, but aimless, as you then must further assume acceleration to be negative (which I didn't).

In which case,

$d = \frac {-9.8x^2}{2} - 20x + 245$

If you've fallen 100ft from 245ft

$245 - 100 = \frac {-9.8x^2}{2} - 20x + 245$

Solve for x and you get how many seconds you've been travelling. Plug that in for v = -9.8x - 20

5. Alternatively in order to solve the second part you can use the conservation of mechanical energy, that is the formula $E=\frac{mv^2}{2}+mgh=\text{a constant}$.

6. Without mass?

7. Originally Posted by derfleurer
Without mass?
Are you talking to me? If so, I included mass (m) in the equation and it cancels out since $E_{\text{initial}}=E_{\text{final}}$.