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How to get R= 2 ( square root ) h1h2

  • #1
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Problem :

A ball is let down a ramp on top of a table with initial velocity of 0 ms-1. When it reaches the end of the ramp, it is launched horizontally. Knowing that we don’t take air resistance or friction into account, and that the height of the ramp is h1, and that of the table is h2 relative to ground level,how do we get to the equation R (range or distance traveled from the end of the ramp) = 2√(h1h2)

Equations needed:

Since it is a projectile motion related exercise, the equations needed are Vv= vcosΘ , Vh= vsinΘ
and all the cinematic equations.

My approach

Since it’s projected horizontally i tried solving it taking into account that Vv = 0 and that Vh= t1h1. So I had to multiply gt1 by time (t2) to obtain the range. And t is equal to v/a or an alternative, t= √(2h2/g). But I wasn’t able to get to the equation R= 2√(h1h2). How do I ?
 
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Answers and Replies

  • #2
PeroK
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A ball is let down a ramp on top of a table with initial velocity of 0 ms-1. When it reaches the end of the ramp, it is launched horizontally. Knowing that we don’t take air resistance or friction into account, and that the height of the ramp is h1, and that of the table is h2 relative to ground level,how do we get to the equation R (range or distance traveled from the end of the ramp) = 2√(h1h2)

Since it is a projectile motion related exercise, the equations needed are Vv= vcosΘ , Vh= vsinΘ
and all the cinematic equations. Since it’s projected horizontally i tried solving it taking into account that Vv = 0 and that Vh= gh1. So I had to multiply gh1 by time (t) to obtain the range. And t is equal to v/a or an alternative, t= √(2h2/g). But I wasn’t able to get to the equation R= 2√(h1h2). How do I ?
How did you get ##v_h = gh_1##?
 
  • #3
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How did you get ##v_h = gh_1##?
I made I mistake, it is meant to be gt, i’ll correct it.
 
  • #4
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I made I mistake, it is meant to be gt, i’ll correct it.
For the motion down the ramp, do the kinematic equations help? Or, is there another way to study this motion?
 
  • #5
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For the motion down the ramp, do the kinematic equations help? Or, is there another way to study this motion?
I guess you could use the kinetic energy, but I used the kinematic equations so I use the equations that I already seen in the book, despite the fact that I can use them.
 
  • #6
PeroK
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I guess you could use the kinetic energy, but I used the kinematic equations so I use the equations that I already seen in the book, despite the fact that I can use them.
So, what do you have for the motion down the ramp?
 
  • #7
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So, what do you have for the motion down the ramp?
since v = u + at and u = o I get that v = at or v = gt. V would be the initial velocity for the projectile motion. I can use this equation since the exercise supposed that there is no friction it is approximately like if it would be free falling. Any thoughts ?
 
  • #8
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since v = u + at and u = o I get that v = at or v = gt. V would be the initial velocity for the projectile motion. I can use this equation since the exercise supposed that there is no friction it is approximately like if it would be free falling. Any thoughts ?
How are you going to find ##t##?

Note, also, you are told nothing about the shape or angle of the ramp.
 
  • #9
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How are you going to find ##t##?

Note, also, you are told nothing about the shape or angle of the ramp.
What the exercise wants me to do is not solve the problem, which would mean knowing the maximum range that the ball will travel. What the exercise wants me to do is to demonstrate that the problem can be solved by using R = 2√(h1h2).
 
  • #10
PeroK
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What the exercise wants me to do is not solve the problem, which would mean knowing the maximum range that the ball will travel. What the exercise wants me to do is to demonstrate that the problem can be solved by using R = 2√(h1h2).
What do you know about energy?
 
  • #11
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What do you know about energy?
The basics. KE= mv2/2 and PE= mgh. And of course ME= KE + PE
 
  • #12
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The basics. KE= mv2/2 and PE= mgh. And of course ME= KE + PE
Okay. Use that for the motion down the ramp.
 
  • #13
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Okay. Use that for the motion down the ramp.
I can’t. Just looked at the exercise again. One of the follow up questions is “ explain why you didn’t need to know the mass of the ball for your answer”.
 
  • #14
PeroK
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I can’t. Just looked at the exercise again. One of the follow up questions is “ explain why you didn’t need to know the mass of the ball for your answer”.
What is this word "can't"? At least try.
 
  • #15
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What is this word "can't"? At least try.
I got the equation right, thank you. Also realized how to accomplish the same thing with kinematics. Thanks for the help
 

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