Robin and the rat

Let's say that Robin aims his arrow exactly at the rat and that the rat doesn't step off. What will happen? Well, since there's gravity the arrow will describe the proverbial parabola and it will hit somewhere on the pillar. Now image the case without gravity; the arrow will continue on a straight line with the initial velocity Robin gave it and finally will hit the rat. So for a given time, what will be the difference between a point on the straight line trajectory and the parabola. Since in both cases there's no acceleration in the $x$ (or horizontal) direction, the motion in this direction will be the same. On the other hand, if we take the difference in the position in the $y$ (or vertical) direction we have:

Straight line: $y = V_0 \sin(\theta) ~t$

Parabola: $y = V_0 \sin(\theta) ~t - \frac{g t^2}{2}$

We see that the difference is the term $\frac{g t^2}{2}$ which is exactly the distance an object falls in a time $t$ if we let it go with no initial velocity. So if Robin aims exactly at the rat, and when he shoots the rat steps off, then in a time $t$ the rat will have fallen exactly the same distance than the arrow in the vertical direction (provided that the initial velocity is large enough to reach the pillar, because otherwise the arrow could just hit the ground in front of the pillar; but we know this is not the case since it says that Robin successfully shoots the rat). So we have a geometry problem: find the angle of a right angled triangle with legs $10 ~m$ and $4.24475 ~m$ , which is: $\tan(\theta) = \frac{4.24475}{10} \Rightarrow \theta \simeq 23°$

We know that everything accelerates at the same rate, so the arrow and the rat must be falling at the same rate. The accelerations due to gravity end up cancelling out because they are both accelerating at $\frac{-9.8 meters}{1second}$ .
Even better, from this, we know that if Robinhood points the arrow straight on top of the pillar (and releases it the same time the rat jumps) it will fall at the same rate as the mouse (killing it when they cross paths).
Drawing a right triangle, it can be observed that $tan\theta = \frac{4.24475}{10}$ .
This means theta is around $\boxed{23 degrees}$ .

This question is incredibly short than it seems to be. Assume the bullet(say) is shot at an angle θ. Instead of observing the act from ground frame, we will use a frame which is accelerating with an acceleration g vertically downwards, the rat will be in rest from this frame, and the bullet will move with constant velocity. This implies that the line along which the bullet is travelling should pass through the apex of the pole where the rat was initially sitting, hence a right angle triangle is complete with height 4.24475 and base 10 with angle b/w hypotenuse and base = θ. Therefore tanθ=(4.24474/10). Hence calculate θ.

Given Height of pillar (Perpendicular) = 4.24475 m Given Distance of Robin (Base) = 10.

tan x = Perpendicular/Base

thus, tan x =4.24475/10 => tan x= 0.424475 => x = tan^-1(0.424475) => x = 23

The pillar is 4.24475 metres high, which is the opposite length. The distance to the pole is 10 metres, our adjacent length. Thus, we take the arctan of 4.24475 divided by 10, giving 23 degrees

10^{2}+4.2275^{2}= AB

AB = considering it as right angled triangle where base = 10 and perpendicular is 4.24475.

AB= 10.86

Resolving AB into two components i.e. AB \cos \theta and AB\ sin \theta .

AB \cos \theta = 10

10.86 \cos \theta=10

\cos \theta=10/10.86

\cos \theta=0.92081

\cos \theta =cos 23 degree

\theta=23degree

angle = arctan(4.24475/10) = 23

If we consider a right angle triangle and height of pillar is its perpendicular and its base is the distance between Robin hood and the pillar then by applying

tan(angle)=Perpendicular/Base

we can find angle

Robin shoots at a point that is $4.24475$ m above the ground that is $10$ m away from where he is standing. Thus, the degree he shot at is approximately $\tan^{-1} \frac{4.24475}{10} = \boxed{23}^{\circ}$ .

Let the ground be the base and the pillar be the perpendicular of a right triangle...

Let's assume the required angle as $\theta$ ... Then...

$\tan \theta = \frac {Perpendicular}{Base} = \frac {4.24475m}{10m} = 0.424475$ $\Longrightarrow \tan \theta = 0.424475 \Longrightarrow \theta = \tan ^{-1} 0.424475=23$

Hence, the required answer is $\fbox {23}$ ...

This is a very famous problem, known as the Monkey and the Hunter . The solution to it is that Robin must aim directly at the rat, because the arrow and the rat fall at the same speed.