Respuesta :
To get the distance, find the unchanging horizontal component of velocity and the time of flight and multiply them.
First find the horizontal component of velocity, Vh by using the equation for centripetal acceleration: v² / r = a Vh² / R = 2g Vh = 2gR
Therefore the horizontal component of velocity is:
►Vh = √( 2gR )
Now to find time elapsed. In the vertical dimension, the ball started its flight with no initial velocity. We determine how far it travels vertically by the equation:
d = ½at²
And substitute our values for d and a and solve for time:
2R = ½gt² t² = 4R / g t = √( 4R / g ) t = 2√( R / g )
►∆t = 2√( R / g )
Now we multiply our two critical values together to get the distance:
d = Vh•∆t
d = [ √( 2gR ) ]•[ 2√( R / g ) ]
d = 2√( 2 R² )
d = 2√2 R
First find the horizontal component of velocity, Vh by using the equation for centripetal acceleration: v² / r = a Vh² / R = 2g Vh = 2gR
Therefore the horizontal component of velocity is:
►Vh = √( 2gR )
Now to find time elapsed. In the vertical dimension, the ball started its flight with no initial velocity. We determine how far it travels vertically by the equation:
d = ½at²
And substitute our values for d and a and solve for time:
2R = ½gt² t² = 4R / g t = √( 4R / g ) t = 2√( R / g )
►∆t = 2√( R / g )
Now we multiply our two critical values together to get the distance:
d = Vh•∆t
d = [ √( 2gR ) ]•[ 2√( R / g ) ]
d = 2√( 2 R² )
d = 2√2 R
The distance from the bottom of the chute that the ball would land is [tex]2\sqrt{2R}[/tex]
Given the following data:
- Centripetal acceleration = 2g.
- Distance, S = 2R.
How to calculate the distance.
In order to calculate the distance from the bottom of the chute that the ball lands, we would determine the time of flight and the horizontal component of velocity.
For the time of flight, we would apply the second equation of motion:
Note: The ball has no initial velocity (u = 0) in the vertical dimension.
[tex]S=ut +\frac{1}{2} gt^2\\\\2R=(0)t +\frac{1}{2} (g)t^2\\\\4R= gt^2\\\\t=\sqrt{\frac{4R}{g} }\\\\t=2\sqrt{\frac{R}{g} }[/tex]
The horizontal component of velocity.
We would determine the horizontal component of velocity by using the formula for centripetal acceleration:
[tex]a = \frac{V^2}{R} \\\\2g = \frac{V^2}{R}\\\\V^2 =2gR\\\\V=\sqrt{2gR}[/tex]
Now, we can determine the distance:
[tex]Distance = Velocity \times time\\\\Distance = \sqrt{2gR} \times 2\sqrt{\frac{R}{g} }\\\\Distance = 2gR \times \frac{4R}{g}\\\\Distance^2 =\frac{8gR^2}{g} \\\\Distance = \sqrt{8R^2} \\\\Distance = 2\sqrt{2R}[/tex]
Read more on centripetal acceleration here: https://brainly.com/question/2788500
Complete Question:
A ball is launched up a semicircular chute in such a way that at the top of the chute, just before it goes into free fall, the ball has a centripetal acceleration of magnitude 2g. Also, the distance between the bottom and the top of the chute is 2R.
How far from the bottom of the chute does the ball land?
