Finding Horizontal Distance in Projectile Motion

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SUMMARY

The problem involves calculating the horizontal distance a 0.5kg rock travels when projected from a height of 6.89m with an initial velocity of 13.7m/s at an angle of 60 degrees. The relevant equations include dy = Voyt + 1/2at^2 for vertical motion and Vx = dx/t for horizontal motion. The solution requires breaking down the initial velocity into its x and y components using Vx = Vcos(θ) and Vy = Vsin(θ). By organizing known variables and applying the equations systematically, the horizontal distance can be accurately determined.

PREREQUISITES
  • Understanding of projectile motion principles
  • Familiarity with trigonometric functions (sine and cosine)
  • Knowledge of kinematic equations for motion
  • Basic algebra for solving equations
NEXT STEPS
  • Study the derivation and application of kinematic equations in projectile motion
  • Learn how to decompose vectors into their components using trigonometric functions
  • Explore the concept of time of flight in projectile motion scenarios
  • Practice solving similar problems involving different angles and initial velocities
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Students studying physics, particularly those focusing on mechanics and projectile motion, as well as educators looking for problem-solving strategies in kinematics.

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Homework Statement


A 0.5kg rock is projected from the edge of the top of a building with an initial velocity of 13.7m/s at an angle 60o above the horizontal. The building is 6.89m in height . At what horizontal distance, x, from the base of the building will the rick strike the ground? Assume the ground is level and that the side of the building is vertical. The acceleration of gravity is 9.8m/s2.


Homework Equations


dy = Voyt + 1/2at^2
Vx = dx/t

The Attempt at a Solution


I have tried to find the vertical and horizontal component of the rock at its highest point. Using the vertical component, I found a time, which would be how long it took for the rock to get to its highest point from its original position. Then, I added the vertical component and the height of the building, and used that as the dy to find the time it takes for the rock to go from its highest point to the ground. I added the two times, and found a dx.

My assignment is a quest on the UTexas site, so it would tell me when my answer is wrong.

And here's a diagram:
untitled-2.jpg


Thank you for your help!
 
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You're making this WAY more complicated than you need to. Did you learn that the x-component of velocity is Vcos \theta and the y component of velocity is Vsin \theta, where theta is the angle made with the x axis?

You're given everything you could possibly want in this problem, just plug and chug.

Edit: To clarify, just write down what you do know. You know y-initial, y-final, x-initial, Vx initial, Vx final, Vy initial, acceleration in the x direction, acceleration in the y direction. The only things you don't know are time, x-final, and Vy-final. You have 3 unknowns and 4 equations (2 for x, 2 for y). Plug and chug.

The way I do every one of these problems is to write down a list of everything you could possibly know about the system. That's 11 items. Initial and final position, initial and final velocity, acceleration, and time. Do that for the x and the y directions. Time is the same, so you only put that down once. After you list those 11 items, fill in what you know, and use that to solve for what you don't.

If you do that, solving for it's highest point, or doing anything fancy like that doesn't even come into play.
 
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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?

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