Darts are Demanding

Classical Mechanics Level 3

At a sports bar, Joe is throwing darts at a dartboard in hopes of hitting the "bullseye," a very small circular portion right at the center of the dartboard.

Joe is standing $6\text{ m}$ from the dartboard, and the bullseye is $5\text{ cm}$ in diameter. Joe throws the dart from exactly the same height as the bullseye, with an initial velocity which makes an angle of $30^\circ$ with the horizontal.

Call the "nominal throwing speed" the speed that would put the dart right in the middle of the bullseye.

If, in general, Joe throws the dart with a speed between 99% and 101% of the nominal throwing speed (distributed uniformly between the extremes), what is the probability (in percent) that Joe will land the dart somewhere within the bullseye?

Details and Assumptions:

The ambient gravity is $10\text{ m/s}^2.$
There is no horizontal (side-to-side) error in Joe's aim.

The answer is 36.09.

1 solution

Riccardo Baldini
Oct 25, 2018

First of all we need to solve a simple projectile motion problem to find the "nominal throwing speed" $v_0$ and impose the passage through the point $(L=6,0)$ :

$x=v_0 \cos\theta t=L$

$y=v_0 \sin\theta t-\frac{g}{2}t^2=0$ $\\->y=f(x,v_0) = x \tan(\theta)-\frac{g x^2}{2 v_0^2\cos^2\theta}$

Which gives (after some calculations): $v_0=\sqrt{\frac{g L}{sin{2\theta}}}$ .

Now we know that the range of speed is between $v_+=101\% v_0$ and $v_-=99\%v_0$ and it's distributed uniformly, so the requested probability is the ratio between the diameter $d=0.05cm$ and the distance $f(L,v_+)-f(L,v_-)$ :

$prob=100\times\frac{d}{f(L,v_+)-f(L,v_-)}=100\times 0.05/(0.1386)=36.1\%$

Darts are Demanding

The answer is 36.09.

1 solution

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