What Is the Kinetic Energy of a Golf Ball at Its Highest Point?

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Homework Help Overview

The discussion revolves around calculating the kinetic energy of a golf ball at its highest point, using the context of projectile motion. The problem involves a golf ball with a specified mass and initial speed, and participants are exploring the implications of these values on kinetic energy and velocity components.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning, Assumption checking

Approaches and Questions Raised

  • Participants are questioning the use of the kinetic energy formula and whether it applies at the highest point of the ball's trajectory. There are discussions about the horizontal and vertical components of velocity and the need for the angle of projection to fully understand the problem.

Discussion Status

Some participants have provided guidance on breaking down the problem into its components, suggesting the use of trigonometric relationships to find the angle of projection and the corresponding velocity components. There is an ongoing exploration of the necessary information and calculations required to address the problem effectively.

Contextual Notes

Participants note that the problem is two-dimensional and emphasize the importance of considering both the horizontal and vertical components of the velocity. The angle of projection is identified as a missing piece of information that is crucial for solving the problem.

shawonna23
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Can someone please tell me what I did wrong for Part A. I used this equation: KE=1/2m*v^2. The answer I got was 58.8J but its wrong. Am I using the right equation?

A 47.0 g golf ball is driven from the tee with an initial speed of 50.0 m/s and rises to a height of 23.4 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 7.0 m below its highest point?
46.7m/s
 
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how fast is the ball going when it is at Hmax?
this is the v you need to use.
 
photon-mass,
how fast is the ball going when it's at Hmax is actually the horizontal velocity
vx = vo x cosa (where vo is the initial velocity and a is the angle of projection)
I think the additional needed info in this question is the angle of projection.

XMLT
 
I hope this site may help you Shawonna
http://www.ac.wwu.edu/~vawter/PhysicsNet/Topics/Vectors/ProjectilesMotion.html

XMLT
 
Last edited by a moderator:
shawonna23 said:
A 47.0 g golf ball is driven from the tee with an initial speed of 50.0 m/s and rises to a height of 23.4 m.
(a) Neglect air resistance and determine the kinetic energy of the ball at its highest point.

Hi,

They are making this a bit confusing for you i think. Keep in mind this is a two dimensional problem, so you need to calculate the x and y components of the velocity v by using the well known triangle-rules with sine and cosine.

At the highest point, the y-component of the velovity is 0 because you do not move upwards anymore. So v_y = 0

The x-component is constant v_x = v' * cos (a) where a is angle of inclination of the ball with respect to the x-axis. The v' is the given initial velocity. What you need to find is the angle a. Try using this t = v'_y/g is the time that the y component of the velocity is 0. This t-value must be substituted into the expression for the y-component of the position (max height) : y - y' = v'_y * t - gt²/2. And y-y' = 23.4. Now you can calculate v'_y and you know v'. Via v'_y = v' * sin (a) you will find a.


marlon
 

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