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of the chute, just before it goes into free fall, the ball has a centripetal acc

ID: 2127447 • Letter: O

Question

of the chute, just before it goes into free fall, the ball has a

centripetal acceleration of magnitude 2 class="mathjax_preview">gstyle="border-left: 0em solid; display: inline-block; overflow: hidden; width: 0px; height: 0.963em; vertical-align: -0.333em;">style="display: inline-block; width: 0px; height: 2.4em;">style="display: inline-block; overflow: hidden; height: 1px; width: 0.003em;">title="g" id="mathjax-span-4" class="mi">title="g" id="mathjax-hitbox-3" class="mathjax_hitbox">id="mathjax-span-3" class="maction">id="mathjax-span-2" class="mrow">style="position: absolute; clip: rect(1.773em, 1000em, 2.79em, -0.544em); top: -2.4em; left: 0em;">style="display: inline-block; position: relative; width: 8px; height: 0px; font-size: 125%;">id="mathjax-span-1" class="math">.

How far from the bottom of the chute does the ball land?


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 2 g. 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

v^2/R = 2g

=> v = sqrt(2gR)

to reach ground in vertical direction:

d = ut+0.5*a*t^2

=> 2*R = 0 + 0.5*g*t^2

=> t = 2*sqrt(R/g)

in horizontal direction:

d = u*t = sqrt(2gR)*2*sqrt(R/g) = 2*R*sqrt(2)


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