# Find the distance that the ball reaches after breaking the rope [closed]

A ball attached to a rope of $$1m$$ of radius, describes circles with a frequence of $$f=10s^{-1}$$ in a horizontal plane at a height of $$3m$$ over the floor. If at a certain moment, the rope breaks... find:

a) The distance that the ball reaches right after the rope is broken.

Attempt. So I've first drawn a diagram but I'm unsure I am right about what is happening. On any case, I tried going with $$\omega=2\pi f\implies \omega=20\pi\ rad/s$$, hence the velocity after it breaks, I'm assuming it is found by $$V=\omega\cdot r=20m/s$$. And from this moment my confusion comes in because I don't know if I should treat the movement after the rope breaks as a projectile motion, but I've tried that and I don't know the angle with the horizontal and I can't seem to be able to get anywhere. I'm stuck

• I don't know the angle with the horizontal. “circles ... in a horizontal plane” Commented Feb 9, 2021 at 21:07
• that made no sense in my head, how can one ball circle in a horizontal plane? Commented Feb 9, 2021 at 21:13
• on any case, I did this to find the the time $t$ in order to later plug that time in the $x=vt$. $$0=3+20\pi t-\frac12 gt^2$$ from the equation $y=y_0+v_0 t-\frac12 gt^2$, but this gives me an approximated value of $t=12.828s$ which when plugging into $x=vt$ doesn't match at all with the answer on my book Commented Feb 9, 2021 at 21:20
• You have misunderstood the description. Like G.Smith says the ball is revolving in the horizontal plane, like a tetherball. Your equation describes the situation when the ball revolves in a vertical plane. Commented Feb 9, 2021 at 21:30
• well then I don't know what to do, I've tried a lot of things (the ones I mentioned), I'd love to be given some good hint Commented Feb 9, 2021 at 21:41

The situation is something like this:

So here according to the question, the radius of the rope is 1m and and height of the circular motion is 3m above the floor.

As you have written about your attempt, you have correctly figured out the angular velocity $$\omega$$ of the ball but the velocity of the ball must be $$v = r\omega$$ = (1$$m$$)*(20$$\pi$$ $$rad/s$$) = 20$$\pi$$ $$m/s$$

Then, after the rope breaks, use the concept of horizontal projectile motion as given in the below figure:

Where your initial height is 3m, the angle of projection is 0 (as the ball is projected from the horizontal plane) and the velocity of the projectile is 20$$\pi$$ m/s.

So by applying the equations:$$x = v_{0x}t$$ and $$y = y_0+v_{0y}t+\frac{1}{2}at^2$$ along the horizontal and vertical directions respectively where $$v_{0y} = 0$$ (As the initial velocity is horizontal. There is no vertical component of the initial velocity), you must get the answer. Hope it helps you.

You can figure horizontal and vertical components separately. How much time will it take for it to fall vertically 3 meters from rest? How far will its linear velocity carry it horizontally in that time?