How Do I Find This? [Physics question]

[SophiaRivera007]

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

[swansont]

What kinematics equations would apply here?

[Country Boy]

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

[Function]

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

[Sriman Dutta]

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

[SophiaRivera007]

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.

[AshBox]

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