Projectile Motion: Finding the Angle of Launch

Click For Summary

Homework Help Overview

The discussion revolves around a projectile motion problem involving a soccer ball kicked from a building. The problem specifies the initial speed, time of flight, and horizontal distance traveled, with the goal of finding the launch angle.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning

Approaches and Questions Raised

  • Participants explore the relationship between speed, distance, and time, and discuss the components of velocity. There are attempts to visualize the problem using a right-angled triangle to relate the components of the initial velocity.

Discussion Status

Some participants have provided guidance on calculating horizontal speed and relating it to the initial speed. There is a progression in the discussion with attempts to derive the angle, though not all participants express confidence in their understanding.

Contextual Notes

There is an indication of confusion among participants regarding the problem setup and the equations needed to solve it. Some participants express feeling lost or unsure about their approach.

helpme2012
Messages
11
Reaction score
0
Help with projectile motion :(

1. Homework Statement
A soccer ball is kicked from the roof of a 260m building. The ball is given an initial speed of 22m/s. It is airborne for a period of 8.3s and travels horizontally for 165.5m before hitting the ground. What is the angle ( in degrees) at which the soccer ball was kicked?


2. Homework Equations
No idea :(

3. The Attempt at a Solution
I don't even know where to begin
 
Last edited by a moderator:
Physics news on Phys.org


You could start with speed= distance /time.
Then think about components of velocity.
 


I honestly have no idea what I'm doing on this question. I'm completely lost.
 


Horizontally the ball moves 165.5m in 8.3s. What is the horizontal speed?
 


19.94m/s
 


19.94 is a component of the initial 22m/s.
Draw a right angled triangle with 22 on the hypotenuse, 19.94 on the adjacent side and work out the angle.
 


24.99 degrees, rounded off to 25 degrees.
 


Yep that's the right answer. That is all that was needed.
 


Hah, I got that before when playing around with Δdχ= VixΔt
You're awesome, I'm an idiot lol. Thanks :)
 
  • #10


Why did you erase the question? This thread might have helped other people.
 
  • #11


HallsofIvy said:
Why did you erase the question? This thread might have helped other people.
I fixed it.
 

Similar threads

Replies
40
Views
3K
  • · Replies 7 ·
Replies
7
Views
4K
Replies
2
Views
3K
  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 4 ·
Replies
4
Views
3K
  • · Replies 22 ·
Replies
22
Views
5K
  • · Replies 4 ·
Replies
4
Views
2K
  • · Replies 18 ·
Replies
18
Views
5K
  • · Replies 15 ·
Replies
15
Views
27K
Replies
5
Views
2K