Kinetic energy of a projectile

[RuthlessTB]

Homework Statement

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?

Homework Equations

K= [itex]1/2[/itex] m v[itex]^{2}[/itex]

The Attempt at a Solution

I couldn't figure out how to get the angle in order to find the kinetic energy.

 

[ap123]

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

 

[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?

 

[ap123]

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

K_f = U_i + K_i
(where U_f is taken as zero)

 

[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?

 

[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 :

(v^2 - vo^2) + (-mgh) = 0

if you make this correction, you should get the same answer.

 

[RuthlessTB]

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