Physics Circular Motion?

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.

Diagram: http://i216.photobucket.com/albums/cc75/Aeds17/Untitled.jpg

Answer 1

first, the centripetal acceleration
2g = v²/r
so it will exit the tube with initial velocity
v = √(2gr)..............(*)
next
the ball is a height 2R when it leaves the chute and its initial downward velocity is zero you have
2r = ½gt²
so it will hit the ground when
t = √(4r/g)..............(**)
next
since the horizontal velocity is constant throughout the flight (neglecting air resistance) the range is given by
x = vt
substituting (*) and (**) gives
x = √(2gr) * √(4r/g) = √[(2gr)(4r/g)] = √(8r²)
so
x = 2r√2
is the range

,.,.,,.,.,.,.,.

Answer 2

How to attack this problem:

1. Use the equation that relates centripetal acceleration, velocity, and R. (Okay, hint: a = v²/R). Solve that equation for "v". In place of "a", use "2g" (because the problem tells you that the centripetal acceleration is 2g).

2. From Step 1, you've figured out an equation for "v", which is the horizontal velocity at which the ball comes out of the chute. Now figure out the time "t" it takes for the ball to hit the ground. (Okay, hint: When the initial _vertical_ velocity is zero (as it is in this case), you can use the equation: height=gt²/2. Solve for "t", and substitute "2R" for "height")

3. Figure out how far the ball travels _horizontally_ in time "t". (Okay, hint: the _horizontal_ speed never changes; that means it is constantly traveling with a horizontal speed of "v" (same "v" as in Step 1). So you can use the equation: distance = vt.)

Answer 3

Hmmmmmm