# How can I find the maximum velocity if I've already found when it occurs?

I've been working on this problem for awhile and got the second half figured out, but I can't seem to get what the maximum actually is.

Here's the question:

A bullet is fired in the air vertically from ground level with an initial velocity 274 m/s. Find the bullet's maximum velocity and maximum height.

I found that a maximum occurs at time t = $\frac{274}{9.8}$ $\approx 27.959$

So I plugged s ($\frac{274}{9.8}$) in and got that the maximum height $\approx$ 3830.408

I tried looking up how to find it and found stuff telling me to do "the first derivative test." I tried using the examples to solve my problem, but can't get the answer.

Can someone please help me figure out how to get the maximum velocity? Thanks!

I can upload a picture of my work if necessary.

## 2 Answers

I think you need some physical reasoning here. Once you fire your bullet vertically, it's only slowed by gravity until it's velocity is zero and then it starts to fall back. So, in absolute value, velocity is maximum in the moment of firing and when it returns to the surface. If you think about the vertical direction as positive, then the maximum occurs precisely in the firing moment. The reason why the derivative test doesn't work here is the following. Mathematically, it's stated as the proposition:

"If $f$ has a relative maximum in an open interval, then $f'$ is zero in this point"

The problem here is that velocity is a decreasing linear function of time, say $v(t) = v_{0} - gt$, where $v_{0}$ is the firing velocity and $g$ is gravity. Also, you're working on a closed interval $[0, t_{max} ]$, where $t_{max}$ is the instant corresponding to maximum height, calculated as above by you. Notice that $v(t)$ has no relative minimum in $(0, t_{max} )$, and that's why the test doesn't work (indeed, it's derivative is a constant). On the other hand, Weierstrass Theorem assures you that the continuous function $v(t)$ attains a maximum in the compact interval $[0, t_{max} ]$. Since you know it's not in the interior, it must be one end-point, and it's clear that it's actually $0$.

• Wow now I feel dumb haha. Thanks man. – Cozen Oct 8 '13 at 1:00

The function that will give you the speed at a specific time is:

$$\vec v_t = \vec v_0 + \vec a \cdot t$$

Because all these vector are parallel to each other, so we can get rid of the them. Note that we need to change the sign in front of the acceleration, because it's in opposite direction of the speed so we have:

$$v_t = v_0 - at$$

We could think of this as a function of $t$ so we have:

$$f(t) = v_0 - at$$

But this is a line, so the slope is constant. Actually:

$$f'(t) = -a$$

So this means that the speed is constantly decreasing, implying that the maximum speed is at the time of the firing.

If you want to get the maximum height, juts set:

$$f(t) = v_0 - at = 0$$

Because at the time when the bullet reaches maximum height it'll have speed of 0 m/s