Gravitational Potential Energy Problem

[VitaX]

Homework Statement

A 0.720 kg snowball is fired from a cliff 8.10 m high. The snowball's initial velocity is 14.2 m/s, directed 30.0° above the horizontal. (a) How much work is done on the snowball by the gravitational force during its flight to the flat ground below the cliff? (b) What is the change in the gravitational potential energy of the snowball-Earth system during the flight? (c) If that gravitational potential energy is taken to be zero at the height of the cliff, what is its value when the snowball reaches the ground?

Homework Equations

Wg = mgyi - mgyf
Change in U = -W

The Attempt at a Solution

a) Wg = .720*9.8*8.10 = 57.1536 J
b) Change in U = -W = -57.1536 J
c) Ug = 0 J (I believe this is correct because since the initial Ug = 0 and the final height is 0, then the overall has to be 0 J)

I'm not sure about my answers because I didn't utilize the speed or angle they gave me that's one reason why I'm posting on here. I'm not sure if my formulas used are correct or incorrect.
My reference height used was at the bottom of the cliff. That equals zero.

 

[PhanthomJay]

Your answers to a and b look OK. For c, the ref height is the top of the 8.10 m cliff; thus at the top of the cliff, using the PE as zero at the cliff height, then at the bottom of the cliff, PE =?

 

[The legend]

i think your c) is wrong...
because the potential energy at the top of cliff is 0 so, at ground(just before hitting) it should be less than 0..

And i also think that there is no need to use the angle given...

EDIT: this is the second time phanthomjay and me gave answers at almost same time..

 

[VitaX]

So part c) is

Wg = .720*9.8*0 -.720*9.8*8.1 = -57.1536 J

Is that what you are saying? If that's so then basically my answer for all parts is the same value just different signs?

 

[The legend]

basically, yes