# Kinetic Energy

1. Aug 12, 2005

### pkossak

I realize this problem is like ridiculously easy, but I just don't understand why you get the answer this way.

A 0.2 kg ball is thrown with an initial velocity of 20 m/s at an angle of 30 degrees with respect to the horizontal. What is the kinetic energy of the ball at the top of it's path?

I don't understand why in order to get the answer, you have to use the horizontal velocity componenent, Vx. Shouldn't you just use the velocity as given, and plug it into .5mv^2? Thanks

2. Aug 12, 2005

### Fermat

KE = ½mv², where v is the velocity of the ball.

After projection, the ball's velocity can be resolved into horizontal and vertical components. So it's KE at any particular time is,

$$KE = \frac{1}{2}mv_H^2 + \frac{1}{2}mv_V^2$$

But at the top of it's flight, $$v_V = 0$$. so only the horizontal velocity needs to be considered.

3. Aug 12, 2005

### pkossak

great explanation

4. Aug 14, 2005

### gunblaze

u must take note that you are projecting the object at an angle of 30 degrees not vertically upwards. Thus, there is velocity also in the horizontal component. Since there is velocity in the horizontal component, there must also be KE in the direction also.

5. Aug 14, 2005

### James R

$$K=(1/2) mv^2$$

Here, v is the speed. That is:

$$v = \sqrt{{v_H}^2 + {v_V}^2}$$

where $v_H$ is the horizontal component of the velocity, and $v_V$ is the vertical component.

Plugging this into the formula for kinetic energy, we get Fermat's result (above):

$$K = (1/2) m ({v_H}^2 + {v_V}^2) = (1/2) m{v_H}^2 + (1/2)m{v_V}^2$$

In this particular example, at the top of the flight, we have:

$$v_H = 20\cos 30\deg$$
$$v_V = 0$$

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