Calculating Shot Put Release Speed | Projectile Motion Homework Problem

  • Thread starter Thread starter Toranc3
  • Start date Start date
  • Tags Tags
    Motion
Click For Summary
SUMMARY

The discussion focuses on calculating the release speed of a shot put, specifically a 7.30 kg object thrown at an angle of 40.0 degrees. The problem references the world record distance of 23.11 m set by Randy Barnes in 1990 and assumes a release height of 2.00 m. The relevant equations for projectile motion are provided, including horizontal and vertical motion equations. The confusion arises from the interpretation of the angle of release, which is clarified as 40 degrees above the horizontal axis.

PREREQUISITES
  • Understanding of projectile motion principles
  • Familiarity with kinematic equations
  • Basic knowledge of trigonometry
  • Ability to solve algebraic equations
NEXT STEPS
  • Study the derivation of projectile motion equations
  • Learn how to calculate initial velocity components using trigonometry
  • Explore the effects of different release angles on projectile distance
  • Practice solving similar physics problems involving projectile motion
USEFUL FOR

This discussion is beneficial for physics students, educators, and anyone interested in understanding the mechanics of projectile motion, particularly in sports contexts like shot put.

Toranc3
Messages
189
Reaction score
0

Homework Statement


In the shot put, a standard track-and-field event, a 7.30 kg object (the shot) is thrown by releasing it at approximately 40.0 degrees over a straight left leg. The world record for distance, set by Randy Barnes in 1990, is 23.11 m. Assuming that Barnes released the shot put at 40.0 degrees from a height of 2.00 m above the ground, with what speed, in m/s, did he release it?


Homework Equations



x-xo=vox*t

y=yo+voy*t+1/2*a*t^(2)

The Attempt at a Solution



Well I was a bit confused with the wording on this sentence:
In the shot put, a standard track-and-field event, a 7.30 kg object (the shot) is thrown by releasing it at approximately 40.0 degrees over a straight left leg.

It says 40 degrees over a straight left leg. What does that actually mean? Thanks!
 
Physics news on Phys.org
I think 40 degrees over a straight left leg just means 40 degrees above the x-axis.
 

Similar threads

Replies
8
Views
2K
  • · Replies 3 ·
Replies
3
Views
2K
Replies
1
Views
4K
  • · Replies 10 ·
Replies
10
Views
6K
Replies
5
Views
3K
  • · Replies 9 ·
Replies
9
Views
3K
Replies
5
Views
5K
  • · Replies 7 ·
Replies
7
Views
3K
  • · Replies 3 ·
Replies
3
Views
2K
  • · Replies 6 ·
Replies
6
Views
5K