Tetherball

we can first assume that the ball should have a velocity to at least reach to the same height (but not hitting) as the top of the pole. So for the sake of simplicity, let's say the velocity of the ball when the ball comes to the same height as the top of the pole $v_1$ .

Then, the key of this problem is the change from circular motion to a projectile motion . At the beginning, since there was a tension at the rope that forces the ball not to go outward the natural length of the rope, the ball was forced to move in a circular motion. However, when the ball goes upward, the tension force starts to decrease because the gravitational force is replacing it for the centripetal force of the ball, and at certain point, when the portion of the gravitational force equals to the centripetal force, the tension force equals to zero and the ball starts to move in a projectile motion. Here is a primitive picture for better understanding. Let's say the angle between the horizontal line at the top of the pole and the rope when the ball starts to move in a projectile motion $\theta$ and the velocity of the ball when it starts to move in a projectile motion $v_2$ .

Then, by the conservation of energy, we can write $\frac {1} {2} mv_1^2 = mgl\sin (\theta) +\frac {1} {2} mv_2^2$

Also, the portion of the gravitational force must equal to the centripetal force, so $mg\sin(\theta)=\frac {mv_2^2} {l}$

By simplifying these two equations, we can get $v_2^2 = gl\sin(\theta) = v_1^2 - 2gl\sin(\theta)$

Then, now let's go to the projectile motion part. The horizontal displacement function of the ball to hit the top of the pole will be $x=v_2\sin(\theta)t = l\cos(\theta)$

Similarly, the vertical displacement will be $y=v_2\cos(\theta)t-\frac {1} {2} gt^2 = -l\sin(\theta)$

then, we can substitute $t$ into $t=\frac {l\cos(\theta)} {v_2\sin(\theta)}$ to the vertical displacement. Also, by changing $v_2^2$ into $gl\sin(\theta)$ , $\frac {\cos^2(\theta)} {\sin(\theta)} (1-\frac {1} {2\sin^2(\theta)}) = -\sin(\theta)$

It is a pity that I was not able to solve this equation nicely, so I used the calculator. Thus, the angle was $\theta = 35.264$ .

Then, we can plug this value into $gl\sin(\theta) = v_1^2 - 2gl\sin(\theta)$ and get the value of $v_1$ . Thus, the velocity value was $v_1=5.8265$ .

Finally, we can find the value of $v_0$ , the velocity of the ball at the bottom of the pole. By the conservation of energy, $\frac {1} {2} mv_0^2 = mgl +\frac {1} {2} mv_1^2$ Then, the answer is $v_0=\boxed {8.550}$

Note: if someone has a better solution without solving such nasty equation above, I would be very pleased.

Let the tension in the string be $T$ . Boundary condition for string going slack at the top of the ball's circular trajectory is $T=0$ . From this point onwards, the ball behaves like a projectile. Thus at the top of its circular trajectory (let it be called point C), making an angle $\theta$ above the horizontal with the top of the pole, where the string goes slack - $T+mg sin\theta= \frac{mv_c^2}{l}$ Since $T=0$ $mg sin\theta=\frac{mv_c^2}{l} \\ v_c^2=glsin\theta~~~~~~~~~~....(1)$

For the ball to strike the top of the pole, the following conditions must be met

$v_csin\theta t=lcos\theta \rightarrow t=\frac{l}{v_ctan\theta}~~~~~~~~...(2)$

Vertical displacement for the ball now behaving as a projectile

$v_ccos\theta t+\frac{1}{2}gt^2= -lsin\theta$

From (2)...

$v_ccos\theta\left (\frac{l}{v_ctan\theta}\right )+\frac{1}{2} g \left (\frac{l}{v_ctan\theta}\right )^2= -lsin\theta \\ \rightarrow v_c^2[cos\theta tan\theta+tan\theta^2sin\theta]=\frac{gl}{2}~~~~~...(3)$

Replacing value of $v_c^2$ from (1) in (3) and cancelling terms, we get

$sin\theta \left ( \cos\theta tan\theta+tan^2\theta sin\theta \right )=\frac {1}{2}\\ \rightarrow 2sin^2\theta cos^2\theta+2sin^4\theta=cos^2\theta~~~~~~~..(4)$

Substituting $sin^2\theta=a$ we can write (4) as-

$2a(1-a)+2a^2=1-a \rightarrow a=1/3 \rightarrow sin\theta=1/ \sqrt{3}\\ v_c^2=gl\sin\theta =gl\frac{1}{\sqrt{3}}~~~~~....(5)$

From conservation of energy we know that

$v_0^2=v_c^2+2g(l+l\sin \theta) = gl\sin\theta+2g(l+l\sin \theta)\\ v_0^2=gl(2+3sin\theta)$

From (5)..

$\boxed{ v_0=\sqrt{gl(2+\sqrt{3})}= 8.55 ms^{-1} }$