Solving Bottle Thrown with Initial Velocity: Find Final Velocities

Click For Summary

Homework Help Overview

The problem involves a bottle thrown with an initial velocity of 4 m/s at a 45-degree angle, and the goal is to find its final horizontal and vertical velocities before it strikes the ocean. The context is within kinematics, specifically projectile motion.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning, Problem interpretation

Approaches and Questions Raised

  • Participants discuss the lack of time information and its impact on solving the problem. Some suggest focusing on the constant horizontal velocity and the vertical component's behavior. Others explore the implications of the initial and final heights being the same and question how to derive the final velocities from the initial conditions.

Discussion Status

The discussion is active, with participants offering insights into the components of the initial velocity and the nature of projectile motion. Some guidance has been provided regarding the use of trigonometry to find the components, and there is recognition of the relationship between the upward and downward motion of the bottle.

Contextual Notes

There is a noted ambiguity regarding the height from which the bottle is thrown relative to the ocean level, which may affect the final vertical velocity calculation. Additionally, the problem does not specify the time of flight, which is a critical factor in solving for the final velocities.

ryryguy
Messages
4
Reaction score
0

Homework Statement


A bottle is thrown with an initial velocity of 4 m/s at 45 degree angle from the horizon. Find its final horizontal and vertical velocities before striking the ocean.


Homework Equations


change x= innitial velocity (t) + a(t squared)
final velocity= innitial velocity + at
d= average velocity (t)


The Attempt at a Solution



I don't know how to solve this equation because all of the formulas include time and the problem doesn't give time.
 
Physics news on Phys.org
I think you are missing how far below the ocean is below from where the bottle is thrown. Also, you know that the x velocity will be constant because there is no acceleration in that direction. So you will simply have to figure out the y component.
 
it says from the horizon, so I guess the innitial and final heights are the same. I have the answer btw, I just don't know how Kaplan got it. here's the answer...2.8 m/s in both the x an y directions.
 
You might want to take a look at http://www.physicsclassroom.com/Class/1DKin/U1L6a.html" . Click the next button at that site and review how these kinematic equations are used.

To get you started, consider:

You don't need to use any of the kinematic equations

You know the initial velocity. It's a vector, right? You know its magnitude and direction.
What's the magnitude of its vertical component? (Hint: this where the 2.8 came from. Think trig).
How about its horizontal component? Will the horizontal component change?
How about the vertical component?

Will the bottle's vertical motion not come to zero at some maximum height and then accelerate back down toward the ocean?

You didn't specify any difference between the height above the ocean when the bottle left the thrower's hand and the ocean level itself. Therefore, if the bottle decelerates to zero from the intial vertical velocity, then it accelerates back down over the same distance that it went up, what must be the final vertical velocity?
 
Last edited by a moderator:
aha, I see the 2.8 now. I had calculated the sq. rt. of 2 / 2 and the only thing I didn't do was multiply that by four. thanks
 

Similar threads

Replies
8
Views
2K
  • · Replies 17 ·
Replies
17
Views
2K
  • · Replies 3 ·
Replies
3
Views
1K
Replies
5
Views
2K
  • · Replies 3 ·
Replies
3
Views
1K
Replies
11
Views
2K
Replies
3
Views
3K
Replies
3
Views
2K
  • · Replies 10 ·
Replies
10
Views
1K
  • · Replies 2 ·
Replies
2
Views
3K