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?

  • 1
    $\begingroup$ 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

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 $t = 4s$.


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!

  • $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.