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Maximum height reached by a ball using work-energy theorem

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ph123
#1
Mar13-07, 05:39 PM
P: 41
A ball is launched with initial speed v from ground level up a frictionless slope. The slope makes an angle theta with the horizontal. Using conservation of energy, find the maximum vertical height hmax to which the ball will climb. Express your answer in terms of v, g, and theta. You may or may not use all of these variables in your answer.



The max height of a body is given by

mgh = 0.5mv^2
gh = 0.5v^2
h = v^2/2g

since the ball is at an angle, and at max height there is zero velocity in the y direction, the only velocity is that in the x-direction, or vcos(theta).

v^2cos(theta)/2g = hmax

but that isn't right. Anyone have any ideas?



3. The attempt at a solution
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Dick
#2
Mar13-07, 05:52 PM
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Are you SURE the angle of the ramp has anything to do with it? Think about kinetic and potential energy. And why do you think there is any x velocity when the ball reaches it's highest point?
zenmaster99
#3
Mar13-07, 05:56 PM
P: 20
Dick beat me to the punch: Energies are scalars---they don't have a direction associated with them. There is no such thing as "kinetic energy in the [tex]\hat x[/tex] direction.

ZM

ph123
#4
Mar13-07, 05:59 PM
P: 41
Maximum height reached by a ball using work-energy theorem

well, why bother giving me an angle if it has nothing to do with the answer? if i threw the ball at 89 degrees off the roof it would attain a greater max height than if i threw it 45 degrees off the top of the building.
Dick
#5
Mar13-07, 06:01 PM
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The problem says, "You may or may not use all of these variables in your answer." Being on a ramp is different from just flying. When the ball stops at the top, it REALLY stops.
ph123
#6
Mar13-07, 06:05 PM
P: 41
v^2/2g, then.

but what if was throwing the ball off the roof of the building. wouldn't the angle be important then?
Dick
#7
Mar13-07, 06:06 PM
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Yes. Very important.
ph123
#8
Mar13-07, 06:08 PM
P: 41
so, then, imagine i do throw the ball off the top then. Would my wrong answer from my first post, v^2cos(theta)/2g, be the max height in that case?
Dick
#9
Mar13-07, 06:18 PM
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Nope. It's not that simple. As zenmaster said, you can't split the kinetic energy into components. You need to split the velocity. And cos is the wrong trig function. The vertical component is given by a sin.
zenmaster99
#10
Mar13-07, 07:44 PM
P: 20
Quote Quote by ph123 View Post
well, why bother giving me an angle if it has nothing to do with the answer? if i threw the ball at 89 degrees off the roof it would attain a greater max height than if i threw it 45 degrees off the top of the building.
That's possibly true, but for the wrong reason.

If we neglect the ramp entirely, and just throw the ball, then the ball never stops in the [tex]\hat x[/tex] direction. It never dumps all of its kinetic energy into potential energy. And that's the clue here: In this problem, the ball can stop at the top of the ramp. Neglecting friction, the only place it can put that kinetic energy is into potential energy.

Depending on how much velocity is lost attaining maximum height, the 45 degree trajectory could go higher than the 89 degree trajectory.

Make sense?

ZM


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