Kinetic/Potential Energy and Work help

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Homework Statement


A baseball outfielder throws a 0.15kg ball at 45m/s at an initial angle of 35°.
a) What is the kinetic energy of the ball as it leaves the player's hand?
b) What is the kinetic energy of the ball at its highest point of trajectory?
c) Gravity is the only force acting on the ball as it moves to the highest point. How much work did gravity do?
d) How high is the ball?
(C and D must use the work energy theorem)

Homework Equations


Kinetic energy = (1/2)mv2
Grav. potential energy = mgΔy
Work = FΔrcosθ

The Attempt at a Solution


I'm pretty confident on parts A and B, C and D are where I'm slightly confused

a) (1/2)mv2 = 0.5 * .15kg * 45m/s2 = 151.9J

b) 0.5 * .15kg * (45cos(35°))2 = 101.9J

c) I'm not sure where to go mathematically from here, because I don't know the max height, and the only equation I can think of requires it.

I assume the work done by gravity is 50J though, because from part a and b, the change in kinetic was -50, and that'd have to change to grav. potential energy.

If C was correct at 50, this is how I'd think to do D.
d) Grav potential(U) = mgΔy
U/mg = Δy = 50/(.15*9.8) = 34m

If someone can check all my work and explain what's wrong, that'd be great! Thank you!
 

Answers and Replies

  • #2
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Welcome to forum
 
Last edited:
  • #3
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Yes you are right
 
  • #4
haruspex
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I assume the work done by gravity is 50J though, because from part a and b, the change in kinetic was -50, and that'd have to change to grav. potential energy.
Work done by a force is ∫F.ds, where all vectors must be measured in the same sense. If the force, gravity, acts down and the distance moved is up, what sign would the work done have?
 

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