Is Work Dependent on Launch Angle?

[DmytriE]

Textbook problem:

A 0.50-kg projectile is given an initial speed of 20m/s at an angle of 37 degrees to a horizontal surface and lands a certain distance (range) from its launch point. How much work is done on the projectile on landing? (Neglect air resistance).

So the solution is -1.0 x 10² J

I found the kinetic energy using K=1/2mv² substituting the known values in for the variables. Low and behold I got 100 J. Knowing that energy is conserved why is it that I did not have to take the angle at which the projectile was shot to find amount of work at landing?

My only explanation for it is that since energy is conserved the path by which it gets there is independent of the path taken. Is my thinking right or on the right track or am I in left field?

 

[Andrew Mason]

T
My only explanation for it is that since energy is conserved the path by which it gets there is independent of the path taken. Is my thinking right or on the right track or am I in left field?

The energy it gives up is its kinetic energy. Kinetic energy is not conserved. Kinetic + potential energy is conserved. But, so long as the potential energy at launch is the same as the potential energy on landing, the kinetic energy it gives up on landing will be the same as the kinetic energy at launch. This only works if the height of launch is the same as the height of landing.

AM

 

[DmytriE]

This only works if the height of launch is the same as the height of landing.

AM

So since the landing place was at the same height as the launch the need to use the angle for calculations was unnecessary, correct?