Help Needed: Solving Golf Ball Kinetic Energy Problem

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Homework Help Overview

The problem involves a golf ball with a mass of 47.0 g that is driven from a tee with an initial speed of 52.0 m/s and rises to a height of 23.2 m. Participants are discussing how to determine the kinetic energy of the ball at its highest point and its speed when it is 9.0 m below that height, while neglecting air resistance.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning, Assumption checking

Approaches and Questions Raised

  • Participants are discussing the kinetic energy at the highest point, questioning the velocity and energy conservation principles. Some suggest calculating potential energy and kinetic energy at different heights, while others raise concerns about the initial velocity components.

Discussion Status

There are various interpretations of the problem, with some participants providing insights into energy conservation and the relationship between kinetic and potential energy. Guidance has been offered regarding the calculations needed for both parts of the question, but no consensus has been reached on the correct approach.

Contextual Notes

Participants are considering the effects of neglecting air resistance and the implications of the ball's trajectory on its velocity components. There is also a mention of the need to clarify assumptions about the direction of the ball's motion.

miranda82
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I have worked out this problem multiple times and i am still getting it wrong, HELP!

I need some help with this, anyone taking a shot is worth it, i would greatly apprieciate

A 47.0 g golf ball is driven from the tee with an initial speed of 52.0 m/s and rises to a height of 23.2 m.
(a) Neglect air resistance and determine the kinetic energy of the ball at its highest point.
J
(b) What is its speed when it is 9.0 m below its highest point?
m/s
it a/a/a/p if anyone can figure it out. Thank you, M.
 
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You should show us what you've done so far. Were you able to determine the components of the velocity at the time when it left the tee?
 
Hello, miranda82!
Well, the answers are:
a)When the ball reachs maximum height(in this case, 23.2 m) its velocity is null(zero), because it breaks instantly to change its direction. Consequently, its kinetic energy is null(zero) too.

b)How there is no air resistance or any friction force, but only weight force(which is resultant force) which is a conservative force, so , the mechanics energy is conservatived. Therefore,
mechanical energy at the highest point:mgH(according to (a))
mechanical energy at a point 9.0 m below to highest point:mgh+(1/2)*mv^2
therefore,
mgh+(1/2)*mv^2 = mgH, is equivalent to [2g(H-h)]^1/2=v, which is the velocity that you're finding.
PS.:H:maximum height
h:height 9.0m below to highest point.
Now, it's only substitute the datas and good calculus!
 
kastarov said:
a)When the ball reachs maximum height(in this case, 23.2 m) its velocity is null(zero), because it breaks instantly to change its direction. Consequently, its kinetic energy is null(zero) too.
Only if it's going straight up.

I haven't done the calculations, but I assume that if the ball had been going straight up, it would have reached a higher altitude than that. You have to use the altitude given to figure out how much of the initial velocity was in the "up" direction and how much was in the "forward" direction.
 
You are given the initial speed and so can calculate the kinetic energy as the ball leaves the tee. You know the height the ball reaches so you can calculate the potential energy (relative to the ground) there. Since, as Fredrik said, energy is conserved, the kinetic energy at the highest point is The kinetic energy on the ground minus the potential energy at the highest point. You can find the speed of the ball at the highest point from that.

For part b, just figure out how high the ball is when it is "9.0 m below 23.2 meters" and do the same as in (a).
 
HallsofIvy said:
Since, as Fredrik said, energy is conserved...
I didn't say that, but I could have said it. :smile: I actually had another solution in mind, but yours is easier.
 

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