Motion in Two Dimensions of marble

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SUMMARY

The discussion focuses on the projectile motion of a marble rolling off a 2.75 m high platform and landing in a 1.50 m wide hole located 2.00 m from the base. The key equations utilized include the kinematic equation for vertical motion, y(t) = y_0 + (v_0y)t + (1/2)(a_y)t^2, and the horizontal motion equation, x(t) = x_0 + (v_0x)t. The calculated time of flight is 0.75 seconds, leading to a horizontal speed of v_0x = 2.86 m/s. The discussion emphasizes the need to determine the vertical component of the initial velocity, v_0y, to fully solve the problem.

PREREQUISITES
  • Understanding of projectile motion principles
  • Familiarity with kinematic equations
  • Knowledge of gravitational acceleration (9.8 m/s²)
  • Basic algebra for solving equations
NEXT STEPS
  • Learn about the conservation of energy in projectile motion
  • Study the derivation and application of kinematic equations
  • Explore the concept of initial velocity components in two-dimensional motion
  • Practice solving similar projectile motion problems with varying parameters
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Students studying physics, particularly those focusing on kinematics and projectile motion, as well as educators seeking to enhance their teaching methods in these topics.

iwonde
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A small marble rolls horizontally with speed v0 off the top of a platform 2.75 m tall and feels no appreciable air resistance. On the level ground 2.00 m from the base of the platform, there is a gaping hole in the ground. The hole is 1.50 m wide. For what range of marble speeds v0 will the marble land in the hole.
 
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Maybe draw a diagram and label the information you know at certain points. What equations do you know that have to do with projectile motion? Explain where you are stuck with this problem. You must show some work, we won't do it for you.
 
marble.jpg


I'm thinking of finding the final velocity, or is that zero?
 
Use the energy concepts (conservation law) in order to obtain the speed at the bottom, and then work it out from there.
 
I haven't learned that yet. I'm supposed use the kinematics equations for this problem.
 
iwonde said:
I haven't learned that yet. I'm supposed use the kinematics equations for this problem.

You can still work it out. You know the marble must fall on 2.75 meters, so it can be inside the hole. Use v(0) as an unknown and work out the equations, probably since you know the distance, you could calculate the time it the marble takes and then find the value of v(0).
 
This is what I calculated:

Formula used: y(t) = y_0 + (v_0y)t + (1/2)(a_y)t^2
0 = 2.75 + 0 - (1/2)(9.8)t^2
t = 0.75s

Then, I plugged time into the equation x(t) = x_0 + (v_0x)t
2 = 0 + (v_0x)(0.7)
v_0x = 2.86m/s

I don't think my answers are reasonable though. Plus, I'm not sure of how to find v_0y.
 

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