Motion in Two Dimensions of marble

AI Thread Summary
A marble rolls off a 2.75 m high platform and must land in a 1.50 m wide hole located 2.00 m from the base. To determine the range of speeds (v0) that allow the marble to land in the hole, kinematic equations for projectile motion are essential. The time of flight is calculated using the vertical motion equation, yielding a time of 0.75 seconds. This time is then used to find the horizontal speed (v0x), resulting in a value of 2.86 m/s. There is uncertainty regarding the calculation of the vertical component of velocity (v0y), indicating a need for further understanding of projectile motion principles.
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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