Equations of Kinematics in 2-D/Projectile Motion

Click For Summary

Homework Help Overview

The discussion revolves around a kinematics problem involving projectile motion, specifically analyzing the trajectory of a tennis ball struck horizontally from a racket.

Discussion Character

  • Exploratory, Conceptual clarification

Approaches and Questions Raised

  • Participants discuss the initial conditions of the problem, with one attempting to apply symmetry in the trajectory and segmenting the motion into parts. Questions arise regarding the interpretation of the ball's height at launch and the setup of the problem.

Discussion Status

Some participants have provided clarifications regarding the maximum height of the ball being at the launch position. There appears to be a productive exchange of ideas, with one participant expressing gratitude for the clarification.

Contextual Notes

There is a mention of a specific numerical answer, which may influence the participants' understanding of the problem, but it is not the focus of the discussion.

Michele Nunes
Messages
42
Reaction score
2

Homework Statement


A tennis ball is struck such that it leaves the racket horizontally with a speed of 28.0 m/s. The ball hits the court at a horizontal distance of 19.6 m from the racket. What is the height of the tennis ball when it leaves the racket?

Homework Equations


v = vo + at
x = 1/2(vo + v)t
x = vot + 1/2(at2)
v2 = vo2 + 2ax

The Attempt at a Solution


I tried to use symmetry in the trajectory but it just kept coming out weird and not making sense, I tried to split up the trajectory into segments, like from the initial position to the max height and then the max height back to the original horizontal position and then that position to the ground, I was trying to find the total vertical displacement for the second half of the trajectory and then the total vertical displacement for the first half and then I was planning on subtracting the two to find the difference of height which would be the answer but my calculations just weren't coming out right at all.
 
Physics news on Phys.org
by the way the answer is 2.40 m
 
Hey,
The ball is released horizontally so the max height of the ball is at the position it was launched. I think you just misunderstood the set up of the problem.
 
  • Like
Likes   Reactions: Michele Nunes
CAF123 said:
Hey,
The ball is released horizontally so the max height of the ball is at the position it was launched. I think you just misunderstood the set up of the problem.
Ohhhhh, okay the universe has aligned, thank you so much!
 
  • Like
Likes   Reactions: CAF123

Similar threads

Replies
40
Views
3K
  • · Replies 21 ·
Replies
21
Views
2K
  • · Replies 38 ·
2
Replies
38
Views
4K
  • · Replies 3 ·
Replies
3
Views
4K
  • · Replies 14 ·
Replies
14
Views
4K
  • · Replies 11 ·
Replies
11
Views
2K
  • · Replies 3 ·
Replies
3
Views
2K
  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 3 ·
Replies
3
Views
2K
  • · Replies 3 ·
Replies
3
Views
3K