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 2 g. how far from the bottom of the chute does the ball land?

Answer 1

We have centripetal acceleration = $v^2/r = 2g$

So, v = $\sqrt{2gr}$

Now by equation of motion we have S= ut +0.5a$t^2$

S =displacement, u = initial velocity, a= acceleration and t = time

Here S =  2r and a = g , u = 0

$2r = 0+\frac{1}{2} *g*t^2$

$t = \sqrt{\frac{4r}{g} }$

Distance traveled in horizontal direction = $\sqrt{2gr}*\sqrt{\frac{4r}{g} }= \sqrt{8r^2} =2r\sqrt{2}$