1. ## Integrals! :)

I have a homework problem that I need help on:

A stone is dropped from the edge of a roof, and hits the ground with a velocity of -175 ft/s. How high (in feet) is the roof?

I know that the acceleration would be -32ft/s due to gravity, so v(t)=-32t+c.
However, since I don't know how long it takes for the stone to hit the ground, I don't know what to use for time, or my constant c. Can you help me to work through this problem? Thanks!

2. Originally Posted by tennisgirl
I have a homework problem that I need help on:

A stone is dropped from the edge of a roof, and hits the ground with a velocity of -175 ft/s. How high (in feet) is the roof?

I know that the acceleration would be -32ft/s due to gravity, so v(t)=-32t+c.
However, since I don't know how long it takes for the stone to hit the ground, I don't know what to use for time, or my constant c. Can you help me to work through this problem? Thanks!
You've got the right start. Let's back up a bit on that velocity equation.

$\frac{dv}{dt} = -32$

$dv = -32~dt$

$\int_{v_0}^v dv = -32 \int_0^t dt$

$v - v_0 = -32t$

$v = -32t + v_0$

So we see that the constant is the velocity of the object at t = 0. (We could have gone straight to this result, but I feel this explanation is more "transparent.")

Once you have v0 you can integrate again:
$\frac{dx}{dt} = -32t + v_0$

and get
$x = x_0 + v_0t - 16t^2$

By the way, there is a much faster way to do this, but it's a more "Physicsy" way of doing things:
$v^2 = v_0^2 + 2a(x - x_0)$

-Dan

3. Originally Posted by tennisgirl
I have a homework problem that I need help on:

A stone is dropped from the edge of a roof, and hits the ground with a velocity of -175 ft/s. How high (in feet) is the roof?

I know that the acceleration would be -32ft/s due to gravity, so v(t)=-32t+c.
However, since I don't know how long it takes for the stone to hit the ground, I don't know what to use for time, or my constant c. Can you help me to work through this problem? Thanks!
Double integrate