Circular motion release to projectile?

Question

A 60 g ball is tied to the end of a 50-cm-long string and swung in a vertical circle. The center of the circle, as shown in Figure, is 150 cm above the floor. The ball is swung at the minimum speed necessary to make it over the top without the string going slack.

If the string is released at the instant the ball is at the top of the loop, where does the ball hit the ground?

Diagram:
http://i216.photobucket.com/albums/cc75/Aeds17/Untitled-1.jpg

Answer 1

> ...the minimum speed necessary to make it over the top without the string going slack.

The things you know are: the net force on the ball (weight + tension); and the radius of the circular motion. So, use equations that relate force with acceleration (F=ma); and acceleration with velocity (for circular motion, a=v²/r). Since those two equations have something in common (namely the "a"), you can combine them.

To put it in more concrete terms:

Net force at top of circle:
Fnet = weight + tension
Fnet = mg + T

But they said the string is just on the verge of going slack. This means the tension is basically zero. So:

Net force at top of circle:
Fnet = mg

So the acceleration at the top of the circle is:
a = Fnet / m
a = (mg) / m
a = g

For circular motion:
a = v² / r

Since we already found that a=g, substitute:

g = v²/r

Solve for "v"
v = sqrt(gr)

Now you essentially have something travelling horizontally at speed "v", under the influence of gravity. Use this equation to find the time "t" it takes to hit the ground:

height = (vy_initial)(t) - gt²/2

But "vy_initial" is the initial _vertical_ velocity, which is zero, so:

height = -gt²/2

Solve that for "t".

Finally, figure out how far something with horizontal speed "v" can go in time "t". That's the simplest equation: distance = vt.

Answer 2

in theory , it shud land at the base of the stand...