Do i need to find the x and y components?

Click For Summary
SUMMARY

The discussion focuses on the physics of projectile motion, specifically analyzing the time a ball remains airborne when launched vertically versus at an angle of 55 degrees. It is established that the time of flight decreases when the launch angle is not vertical. Participants emphasize the need to derive an algebraic expression for the initial vertical velocity at the 55-degree angle, rather than calculating a specific numerical value. The given time of 1.34 seconds is also highlighted as a critical component for further calculations.

PREREQUISITES
  • Understanding of projectile motion principles
  • Knowledge of trigonometric functions related to angles
  • Familiarity with algebraic expressions and manipulations
  • Basic physics concepts of velocity and time of flight
NEXT STEPS
  • Research the equations of motion for projectile trajectories
  • Learn how to derive vertical and horizontal components of velocity
  • Study the impact of launch angles on time of flight
  • Explore the use of trigonometric identities in physics problems
USEFUL FOR

Students studying physics, educators teaching projectile motion, and anyone interested in understanding the dynamics of angled launches in sports or engineering applications.

crism7
Messages
8
Reaction score
1
Homework Statement
j
Relevant Equations
t = 2vi sin theta / g (i'm not sure)
Do i need to find the x and y components??
j
 
Last edited:
Physics news on Phys.org
Welcome to PF. :smile:

You are given how long the ball stays in the air when shot vertically. What is different when it is shot up at an angle of 55 degrees? It will stay in the air a shorter time, but by how much? What is the initial vertical velocity reduced to when the launch angle is 55 degrees?
 
To @crism7 : Note that @berkeman is not expecting you to find a number for the initial vertical velocity. An algebraic expression will be sufficient. Then see how you can work the given 1.34 s into it.
 
  • Like
  • Informative
Likes berkeman and crism7
The book claims the answer is that all the magnitudes are the same because "the gravitational force on the penguin is the same". I'm having trouble understanding this. I thought the buoyant force was equal to the weight of the fluid displaced. Weight depends on mass which depends on density. Therefore, due to the differing densities the buoyant force will be different in each case? Is this incorrect?

Similar threads

Replies
3
Views
1K
Replies
10
Views
3K
Replies
8
Views
1K
  • · Replies 9 ·
Replies
9
Views
1K
  • · Replies 7 ·
Replies
7
Views
2K
  • · Replies 15 ·
Replies
15
Views
2K
Replies
4
Views
1K
  • · Replies 2 ·
Replies
2
Views
2K
  • · Replies 15 ·
Replies
15
Views
2K
Replies
3
Views
2K