A ball is launched up a semicircular chute in such a way that at the top of the

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. 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.

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. 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.

Explanation / Answer

magnitude of centripetal accleration a = 2 g =19.6 we know a = v ^ 2 / R = 19.6 from this Speed v = v[ 19.6 * R ] we know range S = v v[ 2 H /g] we know height H = 2 R So, S = v[ 19.6 R ] v[ 2 * 2R / 9.8 ] = 2.828 R

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