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My first step is to do something with this:

$32 \frac{feet}{sec^2}$.

From browsing through solutions, I know $$\int_0^4 32 \,t \, \mathrm d t $$ will provide the solution for distance.

I also know that $$\frac{d (32\,t^2/2)}{dt} = 32t $$.

Where did $32t^2/2$ come from?

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    $\begingroup$ read the sections in your text with the formula $y=a/2 t^2 $ where $a$ is acceleration. $\endgroup$ – Maesumi Apr 9 '13 at 19:24
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The formula for the distance covered by a body moving at uniform acceleration is given by:

$$s = ut + \dfrac12 at^2$$

Where $s$ is the displacement, $u$ is the initial velocity, $t$ is the time and $a$ is the acceleration. Since the initial velocity is 0, we get:

$$s = \dfrac12 (32)t^2$$

You differentiate this to get the velocity of the ball after time $t$ seconds, since differentiating displacement gives velocity.

But you're looking for 'how far the ball fall', which means you simply take $s = \dfrac12 (32)t^2$ and put $t = 4s$.

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Acceleration is the derivative of velocity which is the derivative of position.

In other words

a = v' = x'' v = x'

where a : acceleration v : velocity x : displacement.

With this relation we can derive all the formulas you need.

Start with acceleration.

v = ∫a dt = a*t + C where t in this case is time and C is an arbitrary Constant.

By setting t = 0 we get C = V_0 where V_0 is initial velocity

Now we have the formula v = a*t + V_0

using the relationship v' = x we can derive a formula for position

x = ∫a*t + V_0 dt = (1/2)a*t^2 + v_0*t + C, setting t=0 we get x = X_0 where X_0 is initial velocity

Therefore we are left with the equations:

x = (1/2)a*t^2 + v_0*t + x_0 and v = a*t + V_0

now you can plug in your values and get the answer!

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  • $\begingroup$ Welcome to MSE! It really helps to format questions and answers using mathJax (see FAQ). It greatly improves readability. Regards $\endgroup$ – Amzoti Apr 10 '13 at 2:18

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