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How Do I Find This? [Physics question]


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#1 SophiaRivera007

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Posted 27 January 2017 - 07:26 AM

If a ball is thrown upward at an angle of 30 with the horizontal and lands on the top edge of a building that is 20 meters away, the top edge is 5 meters above the throwing point, what is the initial speed of the ball in meters/second?

 

Help me with initial steps.


Edited by SophiaRivera007, 27 January 2017 - 07:27 AM.

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#2 swansont

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Posted 27 January 2017 - 10:58 AM

What kinematics equations would apply here?


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#3 HallsofIvy

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Posted 28 January 2017 - 01:56 PM

   Assuming you are not including air resistance (which would make this problem far, far harder) the kinematic equations would be the usual s= (a/2)t^2+ vt+ d where a is the acceleration vector, v is the initial velocity vector, and d is the initial position vector.  Separating x (horizontal) and y (vertical) components and taking the initial speed to be "v" and the intial position to be d= (0, 0), we have x=  v cos(30)t=(\sqrt{3}/2)v t  and y= (-g/2)t^3+ v sin(30)= -4.9t^2+ (0.5)vt where v is the initial speed. 

 

  Since the ball is to end up "20 meters away, the top edge is 5 meters above the throwing point", x= 20 and y= 5.

 

  Solve the two equations (\sqrt{3}/2)v t = 20 and -4.9t^2+ (0.5)vt= 5 for t and v.


Edited by HallsofIvy, 28 January 2017 - 01:57 PM.

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#4 Function

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Posted 28 January 2017 - 04:29 PM

I've always memorized the following equation (so far hasn't ever failed me yet, since basically every variable is in it)

 

y=y_0+x\cdot\tan{\theta}-\frac{g\cdot x^2}{2v_0^2 \cos{\theta}^2}

 

Is it about 40 ms-1?


Edited by Function, 28 January 2017 - 04:34 PM.

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#5 Sriman Dutta

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Posted 29 January 2017 - 06:34 AM

If we draw a diagram the understanding of the problem becomes easy.


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#6 SophiaRivera007

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Posted 31 January 2017 - 12:06 PM

   Assuming you are not including air resistance (which would make this problem far, far harder) the kinematic equations would be the usual s= (a/2)t^2+ vt+ d where a is the acceleration vector, v is the initial velocity vector, and d is the initial position vector.  Separating x (horizontal) and y (vertical) components and taking the initial speed to be "v" and the intial position to be d= (0, 0), we have x=  v cos(30)t=(\sqrt{3}/2)v t  and y= (-g/2)t^3+ v sin(30)= -4.9t^2+ (0.5)vt where v is the initial speed. 

 

  Since the ball is to end up "20 meters away, the top edge is 5 meters above the throwing point", x= 20 and y= 5.

 

  Solve the two equations (\sqrt{3}/2)v t = 20 and -4.9t^2+ (0.5)vt= 5 for t and v.

Thanks For the solution. Is there any online solutions available to find Acceleration unit conversions or force unit conversion tools?

 

I am not from science student. I have to do some basic examples like this to help one of my friend.


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#7 AshBox

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Posted 1 February 2017 - 04:51 AM

Answer : Answer to Homework Help question removed, per the site rules.  :-)

 

To find Acceleration unit conversions or force unit conversion tools - You can visit Advertise-your-site-again-and-you-get-a-suspension.com


Edited by Phi for All, 1 February 2017 - 05:06 AM.
ad spam link removed

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