# Kinetic energy of a projectile

1. Apr 21, 2013

### RuthlessTB

1. The problem statement, all variables and given/known data
A 1kg mass is projected from the edge of a 20m tall building with a velocity of 20 m/s at some unknown angle above the horizontal. What is the kinetic energy (in J) of the mass just before it strikes the floor?

2. Relevant equations
K= $1/2$ m v$^{2}$

3. The attempt at a solution
I couldn't figure out how to get the angle in order to find the kinetic energy.

2. Apr 21, 2013

### ap123

You don't need to use the angle in this problem.
Hint : something is conserved

3. Apr 21, 2013

### RuthlessTB

Oh now I got it..

First
Ki= 0.5 m vi^2
Kf= 0.5 m vf^2
Ui= m g h
Uf= 0
******************
ΔK + ΔF = 0
which will end up
(v^2 - vo^2) + (-mgh) = 0
(v^2 - 400) + (-200) = 0
v^2= 200+400 = 600 m/s

Kf= 0.5 m v^2
Kf= 0.5 (1) (600) = 300 J
******************

Is my solution right?

4. Apr 21, 2013

### ap123

This is the right idea, but there's a mistake in the calculation.
You can make this simpler by not working out vf first - you don't need it.

Kf = Ui + Ki
(where Uf is taken as zero)

5. Apr 21, 2013

### RuthlessTB

Well the final answer will be 400 J

Can I know why this is the proper way? Is it a special case for projectiles or something?
or the way I used is wrong in general?

Last edited: Apr 21, 2013
6. Apr 21, 2013

### ap123

If you had the initial angle of the velocity, you could have used the standard kinematic equations to solve the problem.
Since this wasn't given, and they asked for the final KE suggests that you should use the energy approach.

Your approach was not wrong - you just did more work by calculating the final speed.

In this part, you've missed out the 1/2 * m part for the KE :
if you make this correction, you should get the same answer.

7. Apr 21, 2013

### RuthlessTB

I got it now, I really appreciate your help.. thanks :)