# Law of conservation of energy

1. Dec 30, 2012

### dot123

1. The problem statement, all variables and given/known data
A particle of mass 0.400 kg is shot from P . The particle has an initial velocity vi with the horizontal component of 30 m/s. The particle rises to a maximum height of 20 m above P. Using the law of conservation of energy, determine :
(a) the vertical component of vi
(b) the work done by the gravitational force on the particle during its motion from P to B
(c) the horizontal and the vertical components of the velocity vector when the particle reaches B
P is in height of 60 m.

2. Relevant equations
Ek = -Ep

3. The attempt at a solution
a.)
1/2mv2 = -(m*g*h)
1/2v2 = -(-9,8*20)
v2 = 2*9,8m/s2*20m
v=19,7 m/s¨
b.)
A=m*g*h
A=0,4kg*9,8m/s2*60m
A=294 J
c.)
1/2mv2 =-(m*g*h)
1/2v2=-(9,8*-80)
v2=2*9,8m/s2*80m
v=39,5m/s

I do not know if I should calculate with minus Ep. And then if the acceleration should be with minus in a.) and in c.) the acceleration positive and height negative?

2. Dec 30, 2012

### barryj

I am a bit confused. You say the particle reaches a max heighth of 20 m but use use 60 m in the solution. Seems like part a is correct. The first statement of part b is correct but the second line makes no sense to me.

3. Dec 30, 2012

### dot123

Sorry, it is confusing. The particle is shot from P (P is in the height of 60m above the ground) and then particle rises to height of 20m above P so the particle is in the total height of 80m above the ground and then reaches the ground - B . I hope that I made it clear.

4. Dec 30, 2012

### haruspex

You can choose to measure distance upwards or downwards - entirely up to you, so long as you are consistent. Note that swapping swaps the sign on both acceleration and height, so leaves the sign of PE unchanged.