Definition of Success problems with projectile motion

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SUMMARY

The discussion centers on solving a projectile motion problem involving the calculation of initial velocity (Vo) and time (t) for a thrown object. The scenario describes a throw from a height of 2 meters at a 30-degree angle to land on a desk 8 meters away and 1 meter high. Participants emphasize the importance of using two equations: one for vertical motion and one for horizontal motion, incorporating the range equation and trigonometric functions. The correct approach involves determining the horizontal and vertical components of velocity and solving for time using the equations of motion.

PREREQUISITES
  • Understanding of projectile motion principles
  • Familiarity with kinematic equations
  • Knowledge of trigonometric functions (sine and cosine)
  • Ability to manipulate algebraic equations
NEXT STEPS
  • Study the derivation and application of the range equation in projectile motion
  • Learn how to decompose velocity into horizontal and vertical components
  • Explore the use of kinematic equations for solving projectile motion problems
  • Practice solving similar projectile motion problems with varying angles and distances
USEFUL FOR

Students studying physics, educators teaching projectile motion concepts, and anyone looking to improve their problem-solving skills in kinematics.

Minestra
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So I just got beat up by this question on my midterm. I'm not sure if these problems are always called definition of success but that is how my professor refers to them as.

The question: (paraphrased)

When you walk into your dorm room you like to throw your keys onto the center of your desk. The center of your desk is 8 meters from you and is 1 meter high. You're nearly 2 meters tall thus your throw well start at 2 meters. You throw it at an angle of 30 degrees. Because of all your practice it lands dead center on your desk. What is the initial Velocity, Vo, of your throw, and at what time, t, does it land?

My attempts and thoughts:

I was at a total loss for this problem, I tried reworking the three constant acceleration equations and got no where. Upon leaving the lecture hall, it dawned on me that the range equation may help but I'm not 100% on that either. Thanks for your help I just want to know how to approach this type of problem in the future.
 
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Write two equations, one for height y above the floor and one for horizontal distance x from where you were standing.

The equations will involve ##x,y,t,V_0## as well as sines and cosines of the known angle 30 degrees. At the landing point we have ##x=8,y=1##, which leaves us with two unknown variables and two equations, hence that can be solved.
 
You are looking to find the velocity required to start at [0,0] with initial angle of 30 degrees and end at [8,-1].
Your horizontal component of velocity is ... ?
Your vertical component of velocity is ... ?
Using vertical component of velocity ##v_{vert},## you can determine the time it takes to return back to y = -1.
## r(t) = v_{vert}*t - 9.8 \frac{t^2}{2} = -1##
Take the positive option for t in terms of v_0.
Using the horizontal component ## v_{horiz},## you can set ##r(t) = 8 = v_{horiz}*t##.
This will give you an equation in terms of the components of v.
The form you get should look just like the range equation, except some forms of the range equation assume that you will end at y=0. If that were the case, you simply shift your starting points accordingly.
 
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andrewkirk said:
Write two equations, one for height y above the floor and one for horizontal distance x from where you were standing.

The equations will involve ##x,y,t,V_0## as well as sines and cosines of the known angle 30 degrees. At the landing point we have ##x=8,y=1##, which leaves us with two unknown variables and two equations, hence that can be solved.
Well there's 20 points I lost by writing Vox and Voy oppose to VoCOS(Theta) etc. Thanks for the response lesson learned.
 
You were not wrong to use ##V_{0x}, V_{0y}##, but the solution hinges upon how they are related to each other by the launch angle.
 
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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