# How Do I Find This? [Physics question]

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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

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swansont    6167

What kinematics equations would apply here?

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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

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Function    76

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

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

Edited by Function

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

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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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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