What Is the Direction of the Ball's Projection?

  • Thread starter Thread starter grando69
  • Start date Start date
  • Tags Tags
    Ball Direction
Click For Summary
SUMMARY

The discussion focuses on determining the direction of a ball's projection when thrown from a height of 5 meters with an initial speed of 20 m/s. The ball impacts the ground after 1.5 seconds, allowing for the application of the equations of motion: s = ut + 1/2at², v = u + at, and v² = u² + 2as. By breaking down the initial speed into horizontal and vertical components, the horizontal component is calculated as ut cos θ, while the vertical component is determined as ut sin θ - (gt²)/2, where g represents gravitational acceleration.

PREREQUISITES
  • Understanding of basic physics concepts, specifically projectile motion.
  • Familiarity with the equations of motion in physics.
  • Knowledge of trigonometric functions and their application in resolving vector components.
  • Ability to perform calculations involving gravitational acceleration (g = 9.81 m/s²).
NEXT STEPS
  • Study the derivation and application of projectile motion equations in various scenarios.
  • Learn how to resolve vectors into components using trigonometric identities.
  • Explore the effects of air resistance on projectile motion and how it alters the equations.
  • Investigate real-world applications of projectile motion in sports and engineering.
USEFUL FOR

This discussion is beneficial for physics students, educators, and anyone interested in understanding the principles of projectile motion and its calculations.

grando69
Messages
1
Reaction score
0

Homework Statement


A child throws a ball with a speed 20ms-1 from a window 5m above horizontal ground. If the ball hits the ground 1.5 seconds later, find the direction of the projection.


Homework Equations


The basic equations of motion:
s=ut+1/2at^2
v=u+at
v^2=u^2+2as


The Attempt at a Solution

 
Physics news on Phys.org
What have you tried? Show your thoughts.
 
try breaking the speed of the ball into component form, so that you have a horizontal and a vertical component. assuming that there is no air resistance, the horizontal component will remain constant as there is no acceleration occurring in the horizontal plane. The only acceleration present is gravity which acts downwards in the vertical plane.
the horizontal component will be: [tex]ut\cos \theta[/tex]
the vertical component will be: [tex]ut\sin \theta - \frac{{gt^2 }}{2}[/tex]
 

Similar threads

Replies
12
Views
1K
  • · Replies 38 ·
2
Replies
38
Views
4K
Replies
34
Views
3K
Replies
7
Views
3K
  • · Replies 4 ·
Replies
4
Views
2K
  • · Replies 9 ·
Replies
9
Views
3K
  • · Replies 3 ·
Replies
3
Views
1K
  • · Replies 4 ·
Replies
4
Views
2K
  • · Replies 4 ·
Replies
4
Views
3K
  • · Replies 7 ·
Replies
7
Views
2K