SophiaRivera007 Posted January 27, 2017 Share Posted January 27, 2017 (edited) 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 January 27, 2017 by SophiaRivera007 Link to comment Share on other sites More sharing options...

swansont Posted January 27, 2017 Share Posted January 27, 2017 What kinematics equations would apply here? Link to comment Share on other sites More sharing options...

Country Boy Posted January 28, 2017 Share Posted January 28, 2017 (edited) Assuming you are not including air resistance (which would make this problem far, far harder) the kinematic equations would be the usual [math]s= (a/2)t^2+ vt+ d[/math] 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 [math]x= v cos(30)t=(\sqrt{3}/2)v t [/math] and [math]y= (-g/2)t^3+ v sin(30)= -4.9t^2+ (0.5)vt[/math] 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 [math](\sqrt{3}/2)v t = 20[/math] and [math]-4.9t^2+ (0.5)vt= 5[/math] for t and v. Edited January 28, 2017 by Country Boy Link to comment Share on other sites More sharing options...

Function Posted January 28, 2017 Share Posted January 28, 2017 (edited) I've always memorized the following equation (so far hasn't ever failed me yet, since basically every variable is in it) [math]y=y_0+x\cdot\tan{\theta}-\frac{g\cdot x^2}{2v_0^2 \cos{\theta}^2}[/math] Is it about 40 ms^{-1}? Edited January 28, 2017 by Function Link to comment Share on other sites More sharing options...

Sriman Dutta Posted January 29, 2017 Share Posted January 29, 2017 If we draw a diagram the understanding of the problem becomes easy. Link to comment Share on other sites More sharing options...

SophiaRivera007 Posted January 31, 2017 Author Share Posted January 31, 2017 Assuming you are not including air resistance (which would make this problem far, far harder) the kinematic equations would be the usual [math]s= (a/2)t^2+ vt+ d[/math] 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 [math]x= v cos(30)t=(\sqrt{3}/2)v t [/math] and [math]y= (-g/2)t^3+ v sin(30)= -4.9t^2+ (0.5)vt[/math] 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 [math](\sqrt{3}/2)v t = 20[/math] and [math]-4.9t^2+ (0.5)vt= 5[/math] 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. Link to comment Share on other sites More sharing options...

AshBox Posted February 1, 2017 Share Posted February 1, 2017 (edited) 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 February 1, 2017 by Phi for All ad spam link removed Link to comment Share on other sites More sharing options...

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