# Mechanics (speed)

1. May 9, 2005

### Kahsi

Hi.

A man is throwing a plate in the air. The plate reaches the height of 40m and the length of 70m. What was the speed of the plate?

This is what I've done

$$v_y:$$

$$mgh = \frac{mv^2}{2} => v = \sqrt{2gh} = \sqrt{2*9,82*40} = 28m/s$$

$$v_x:$$

$$v = v_0 + at => t = 28/9,82 = 2,85s$$

$$s = vt => v = s/t = 70/2,85 = 24,56m/s$$

Code (Text):
/|
/  |
/    | v_y = 28m/s
v_z   /      |
/        |
/          |
/            |
/______________|
v_x = 24,6m/s
$$v_z = \sqrt{v_y^2+v_x^2} = \sqrt{24,6^2 + 28^2} = 37m/s$$

But the answer should be 31m/s. What am I doing wrong?

Thank you.

2. May 9, 2005

### krab

Initial v_y is 28 m/s, but final is -28m/s, so total change is 56m/s and you've calculated t incorrectly; it is actually twice the value you gave.

3. May 10, 2005

### Kahsi

Thank you for the reply krab.

I have now another question.

We know that
$$mgh = \frac{mv^2}{2} => h = \frac{v^2}{2g}$$
And we also know that $$s = v_0t + \frac{at^2}{2}$$
When it's a free fall $$v_0 = 0$$. That gives us
$$s = -\frac{gt^2}{2}$$

So what's the difference between s and h?

Thank you.

Last edited: May 10, 2005
4. May 10, 2005

### James R

Your expression for h is in terms of the speed v when the object hits the ground.

The object's speed at time t is

v = gt

so t=v/g

Putting this into your expression for s, we get:

$$s = -\frac{gt^2}{2} = -\frac{g}{2}\frac{v^2}{g^2} = -\frac{v^2}{2g} = -h$$

So, as you can see, h and s are the same, apart from the negative sign. The negative sign comes from the fact that in one case you've measured positive distances downwards, and in the other positive distances are upwards.