Maximum height reached by a ball using work-energy theorem

In summary: So if we throw the ball off the roof without taking into consideration the ramp, then the angle wouldn't matter and the answer would be the same.If we throw the ball off the roof without taking into consideration the ramp, then the angle wouldn't matter and the answer would be the same.
  • #1
ph123
41
0
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?



The Attempt at a Solution

 
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  • #2
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?
 
  • #3
Scalars and Vectors

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
 
  • #4
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.
 
  • #5
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.
 
  • #6
v^2/2g, then.

but what if was throwing the ball off the roof of the building. wouldn't the angle be important then?
 
  • #7
Yes. Very important.
 
  • #8
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?
 
  • #9
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.
 
  • #10
ph123 said:
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
 

What is the work-energy theorem?

The work-energy theorem states that the work done on an object is equal to the change in its kinetic energy. In other words, the net work done on an object will result in a change in its speed or direction of motion.

How is the work-energy theorem related to the maximum height reached by a ball?

The work-energy theorem can be applied to the motion of a ball to determine the maximum height it can reach. This is because the work done on the ball by the force of gravity will result in a change in its kinetic energy, which in turn, determines its maximum height.

What factors affect the maximum height reached by a ball?

The maximum height reached by a ball is affected by several factors, including the initial velocity of the ball, the angle at which it is thrown, and the force of gravity. These factors, along with the mass of the ball, determine the work done on the ball and therefore its maximum height.

Can the work-energy theorem be used to accurately predict the maximum height reached by a ball?

Yes, the work-energy theorem is a reliable and accurate way to determine the maximum height reached by a ball. However, it is important to note that this theorem assumes ideal conditions and does not take into account air resistance or other external forces.

What are some real-world applications of the work-energy theorem in studying the motion of objects?

The work-energy theorem is used in a variety of fields, such as engineering, physics, and sports. It can be used to analyze the motion of projectiles, determine the efficiency of machines, and understand the dynamics of sports such as basketball and tennis.

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