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# How far from the bottom of the chute does the ball land? your answer for the distance the ball travels from the end of the chute should contain r?

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To get the distance, find the unchanging horizontal component of the velocity and the time-of-flight and multiply them.In the first place, find the horizontal component of the speed Vh by using the equation for centripetal acceleration: v2 / r = a Vh2 / R = 2g V = 2g therefore, the horizontal component of the velocity is: ►Vh = √( 2gr ) Now to find the elapsed time. In the vertical dimension, the ball began its flight without initial velocity. We can determine how far it moves 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 to obtain the distance: d = V•∆t d = [ √( 2gr ) ]•[ 2√( R / g ) ] d = 2√( 2 R2 ) d = 2√2 R