# 1-D Projectile Motion: Throwing a Rock

1. Jan 26, 2017

### Jarvis88

1. The problem statement, all variables and given/known data

You throw a small rock straight up from the edge of a highway bridge that crosses a river. The rock passes you on its way down, 5.00 s after it was thrown. What is the speed of the rock just before it reaches the water 21.0 m below the point where the rock left your hand? Ignore air resistance.

2. Relevant equations

y= y0+v0t-1/2*gt2
v= v0 -1/2*gt.

3. The attempt at a solution
I used v= v0 -1/2*gt to arrive at the velocity that the rock was going at the moment it was passing the hand. My answer was v= 0- 1/2*(9.8m/s2), which gave me 24.5m/s. I have an idea of where to go from here. I know I need to use v2 − 2g(y − y 0).

My issue is that I don't want to just memorize the equation. I need help deriving it from the two basic equations: y= y0+v0t-1/2*gt2 and v= v0 -1/2*gt. Once I do that, I believe I can finish the problem on my own.

2. Jan 26, 2017

### PeroK

So, you want to derive?

$v^2 - u^2 = 2as$

3. Jan 26, 2017

### Jarvis88

I think so, although I'm not sure what the s and u are for.

4. Jan 26, 2017

### PeroK

$u$ is the initial velocity and $s$ is the displacement. I'm not sure why $s$ is used, but it's fairly standard.

5. Jan 26, 2017

### PeroK

Here's my favourite derivation, as it reflects the way I think about kinematic problems:

$v^2-u^2 = (v-u)(v+u)$

Now:

$v-u = at$

Because the difference between the final and initial velocities is just the acceleration multiplied by the time.

And:

$\frac{v+u}{2} = v_{avg} \ \$ is the average velocity (when you have constant acceleration). Hence:

$v+u = 2v_{avg}$

And

$s = v_{avg}t \ \$ because the displacement is simply the average velocity multiplied by the time.

So, putting this all together:

$(v-u)(v+u) = at(v+u) = 2atv_{avg} = 2as$

6. Jan 26, 2017

### PeroK

You need to check where you got this second equation from.