# Thread: Integration and initial velocity

1. ## Integration and initial velocity

With what initial velocity must an object be thrown upward (from ground level) to reach a maximum height of 550 ft?

Use a(t)= -32ft/sec/sec as the acceleration due to gravity.

Could someone help me with this problem? We just began integration and I dont really understand it yet.

2. Have you tried using $v^2=u^2+2as$?

It seems like that kind of problem.

3. You can look at the problem in terms of three functions: the distance $s(t)$, the velocity $v(t)$, and the acceleration $a(t)$. The acceleration has already been given:

$a(t) = -32\frac{\mbox{ft}}{\mbox{sec}^2}$.

Since differentiating velocity gives acceleration, we can go backward from acceleration to velocity by integrating:

$v(t) = \int a(t)\, dt = \int -32\frac{\mbox{ft}}{\mbox{sec}^2}\,dt = -32t\frac{\mbox{ft}}{\mbox{sec}^2} + v_0$.

We put in the constant $v_0$ when integrating to account for the fact that many different functions will have the same derivative -- in this case, $a(t)$. The constant $v_0$ here happens to correspond to the initial velocity with which you would choose to throw the object. (It doesn't matter to gravity, of course, which is why we get the same derivative!)

Now we can integrate $v(t)$ to find the distance $s(t)$:

$s(t) = \int v(t)\, dt = \int (-32t\frac{\mbox{ft}}{\mbox{sec}^2} + v_0)\, dt = -16t^2\frac{\mbox{ft}}{\mbox{sec}^2} + v_0t + s_0$.

Since the initial distance is 0, the constant $s_0$ will be 0:

$s(t) = -16t^2\frac{\mbox{ft}}{\mbox{sec}^2} + v_0t$.

Now, we can use the fact that the maximum height of the function is a point at which

$v(t) = 0$