Finding minimum velocity.

  • Thread starter Miike012
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  • #1
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What is the minimum speed the rock must have, when it leaves the sling to travel exactly 400 m?

The only equation I can think of using is...

y = (tan(theta initial))x - (g/(2Vx0^2)x^2

y = 0
x = 400
However I have to many unknowns... I need to solve for V but I dont know Vx0 or Vy0 or the initial angle...

Help would be appreciated... thank you.
 

Answers and Replies

  • #2
798
1
You should post the entire question as it is relatively difficult to help with what you have given.
 
  • #3
798
1
If that is all you are given, note that
[tex]v_{min}[/tex] will occur when [tex]\theta =\frac{\pi}{4}[/tex]
 
  • #4
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That was all I was given
 
  • #5
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If that is all you are given, note that
[tex]v_{min}[/tex] will occur when [tex]\theta =\frac{\pi}{4}[/tex]
Why will this be a 45 deg angle?
 
  • #6
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Ok I got the answer correct.. but can some one please tell me why initial theta is 45 deg?
 
  • #7
798
1
Proof:
Use the range equation,
[tex]R = \frac{V_{i}^2 sin 2\theta}{g}[/tex]
Solve this equation for the initial velocity. Then treat it like a max/min problem and differentiate the equation, set equal to zero and solve. You will determine that the minimum occurs at 45 degrees.
 
  • #8
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interesting... I probably dont know that because I am not in a calc based phy
 
  • #9
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is that the only way to determine the 45 deg angle by differentiating?
 
  • #10
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and how can the range equation be used for this type of problem? The range equation I thought only deals with situations when a projectile travels some horizontal distance before falling to its original position... but in the case of the problem I posted the object would have fallen to its original position, that being the ground at y = 0.
 
  • #11
798
1
I merely presented the range equation as a method for proving that the minimum occurs at 45 degrees. Moreover, now that you know the angle is 45 degrees, you can plug and chug into any equations that apply for this problem.
 
  • #12
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Proof:
Use the range equation,
[tex]R = \frac{V_{i}^2 sin 2\theta}{g}[/tex]
Solve this equation for the initial velocity. Then treat it like a max/min problem and differentiate the equation, set equal to zero and solve. You will determine that the minimum occurs at 45 degrees.
do i differentiate Initial V with respect to initial theta and keep R constant?
 
  • #14
Redbelly98
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is that the only way to determine the 45 deg angle by differentiating?
No, it's not the only way. You (should) know that sin(anything) has a maximum value of 1, which happens when "anything" is equal to ____?
 

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