Projectile Motion of a grasshopper off a cliff

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SUMMARY

The discussion focuses on calculating the height of a cliff from which a grasshopper leaps, using projectile motion principles. The initial speed of the grasshopper is determined to be 1.5 m/s. Participants emphasize the importance of separating horizontal and vertical motions, applying the equations of motion accordingly. The final solution involves using the time of flight derived from horizontal motion to calculate the vertical distance using the y-component of the velocity.

PREREQUISITES
  • Understanding of projectile motion concepts
  • Familiarity with kinematic equations
  • Knowledge of trigonometric functions (sine and cosine)
  • Ability to solve systems of equations
NEXT STEPS
  • Study the derivation and application of kinematic equations in projectile motion
  • Learn how to decompose vectors into their horizontal and vertical components
  • Explore the concept of time of flight in projectile motion scenarios
  • Practice solving problems involving projectile motion with varying angles and initial velocities
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Students studying physics, educators teaching projectile motion, and anyone interested in applying kinematic principles to real-world scenarios.

Todd88
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Homework Statement



A grasshopper leaps into the air from the edge of a vertical cliff, as shown in the figure below.

1012348.jpg



Homework Equations



Use information from the figure to find the height of the cliff.

The Attempt at a Solution



The first question was to find the initial speed of the grasshopper. I found this to be 1.5m/s.

I have tried using this equation to get the height of the cliff but it isn't right:

h = ut + (1/2)gt2 where u is the initial velocity. Any help is greatly appreciated.
 
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How did you get the 1.5 m/s?
You must separate the vertical and horizontal motion - they are independent.
Make two headings "horizontal" and "vertical". Write x = vt under the horizontal heading and the accelerated motion formulas under the vertical heading. Fill in all the numbers you have. Note that while you don't know the initial Vx or Vy, you do know their ratio, so enter Vx as an unknown and then Vy as Vx times a tangent. If you have only one unknown in one of the formulas, begin with that! If not, you'll have to use 2 or 3 of them and solve as a system of equations.

Show your equations here if you would like more help!
 
Well to get the components of the velocity it is just 1.5sin(50) and 1.5cos(50) so can I just use the y-component and plug that into the formula to find the total distance traveled in the y direction?
 
Okay, I found the 1.5 too. Tricky calc!
I think you will have to use the horizontal distance first - it matters. Use it to find the time of flight and then use the vertical part to find the vertical distance.
 
Okay I got the total time to be 1.0993 s. I used this :

t = 1.06 m / (1.5cos 50)

Not sure where to go to from here...
 
Bah, I got it. Just use that time, then the y-component and use the equation:

x = x0 + V0xt + 1/2axt^2

Simple...Thanks a lot for your help!
 
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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