1) The maximum distance can be reached when the dart is shot with an angle of
![\theta=45^{\circ}](https://tex.z-dn.net/?f=%5Ctheta%3D45%5E%7B%5Ccirc%7D)
above the horizontal (see demonstration of this fact at the end).
2) The motion of the dart is an uniform motion on the x-axis (horizontal direction) with constant velocity
![v_i = 5.84~m/s](https://tex.z-dn.net/?f=v_i%20%3D%205.84~m%2Fs)
and it is an uniformly accelerated motion on the y-axis (vertical direction), with the gravitational acceleration
![g=9.81~m/s^2](https://tex.z-dn.net/?f=g%3D9.81~m%2Fs%5E2)
acting downwards. So we can write the laws of motion on both directions:
![S_x(t) =( v_i \cos{\theta}) t](https://tex.z-dn.net/?f=S_x%28t%29%20%3D%28%20v_i%20%5Ccos%7B%5Ctheta%7D%29%20t)
![S_y(t) = (v_i \sin{\theta})t - \frac{1}{2} gt^2](https://tex.z-dn.net/?f=S_y%28t%29%20%3D%20%28v_i%20%5Csin%7B%5Ctheta%7D%29t%20-%20%5Cfrac%7B1%7D%7B2%7D%20gt%5E2)
where the negative sign means that g points downwards.
3) First of all we can find the time at which the dart reaches the ground. This can be found by requiring
![S_y(t)=0](https://tex.z-dn.net/?f=S_y%28t%29%3D0)
:
![(v_i \sin{\theta})t - \frac{1}{2} gt^2=0](https://tex.z-dn.net/?f=%20%28v_i%20%5Csin%7B%5Ctheta%7D%29t%20-%20%5Cfrac%7B1%7D%7B2%7D%20gt%5E2%3D0)
From this we find two solutions:
![t=0~s](https://tex.z-dn.net/?f=t%3D0~s)
, corresponding to the beginning of the motion (so we are not interested in this one), and
![t= \frac{2 v_i \sin{\theta}}{g} =0.84~s](https://tex.z-dn.net/?f=t%3D%20%5Cfrac%7B2%20v_i%20%5Csin%7B%5Ctheta%7D%7D%7Bg%7D%20%3D0.84~s)
4) Now that we now when the dart reaches the ground, we can use this information to find the distance covered on the x-axis, by using
![t=0.84~s](https://tex.z-dn.net/?f=t%3D0.84~s)
inside the equation of
![S_x(t)](https://tex.z-dn.net/?f=S_x%28t%29)
written at point 2:
![S_x(0.84~s)= 5.84~m/s \cdot \cos{45^{\circ}} \cdot 0.84~s }=3.47~m](https://tex.z-dn.net/?f=S_x%280.84~s%29%3D%205.84~m%2Fs%20%5Ccdot%20%5Ccos%7B45%5E%7B%5Ccirc%7D%7D%20%5Ccdot%200.84~s%20%7D%3D3.47~m)
--------------------------
DEMONSTRATION OF POINT 1:
Calling v the initial velocity of the dart, and using a coordinate system where the x-axis coincides with the horizontal direction and the y-axis with the vertical direction, we can write the law of motion on both directions:
![S_x(t) = v cos{\theta} t](https://tex.z-dn.net/?f=S_x%28t%29%20%3D%20v%20cos%7B%5Ctheta%7D%20t)
![S_y(t) = v sin{\theta} t + \frac{1}{2} g t^2](https://tex.z-dn.net/?f=S_y%28t%29%20%3D%20v%20sin%7B%5Ctheta%7D%20t%20%2B%20%5Cfrac%7B1%7D%7B2%7D%20g%20t%5E2)
where t is the time,
![\theta](https://tex.z-dn.net/?f=%5Ctheta)
is the initial angle of the dart, and
![g= -9.81 m/s^2](https://tex.z-dn.net/?f=g%3D%20-9.81%20m%2Fs%5E2)
is the gravitational acceleration.
The maximum horizontal distance can be found by requiring that
![S_y=0](https://tex.z-dn.net/?f=S_y%3D0)
. This condition occurs twice: when the motion starts (
![t=0](https://tex.z-dn.net/?f=t%3D0)
) and when the dart falls to the ground (let’s call
![t_s](https://tex.z-dn.net/?f=t_s)
the time at which this happens). Therefore we can find
![t_s](https://tex.z-dn.net/?f=t_s)
by requiring
![S_y(t_s)=0](https://tex.z-dn.net/?f=S_y%28t_s%29%3D0)
, i.e.:
![v sin{\theta} t_s + \frac{1}{2} g t_s^2 =0](https://tex.z-dn.net/?f=v%20sin%7B%5Ctheta%7D%20t_s%20%2B%20%5Cfrac%7B1%7D%7B2%7D%20g%20t_s%5E2%20%3D0)
which has two solutions:
![t_s=0](https://tex.z-dn.net/?f=t_s%3D0)
(beginning of the motion) and
![t_s = - \frac{2 v sin{\theta}}{g}](https://tex.z-dn.net/?f=t_s%20%3D%20-%20%5Cfrac%7B2%20v%20sin%7B%5Ctheta%7D%7D%7Bg%7D)
.So we can find the maximum horizontal distance covered by the dart by substituting this
![t_s](https://tex.z-dn.net/?f=t_s)
into the law of motion for
![S_x(t)](https://tex.z-dn.net/?f=S_x%28t%29)
:
Since
![sin2\theta = 2 cos\theta sin\theta](https://tex.z-dn.net/?f=sin2%5Ctheta%20%3D%202%20cos%5Ctheta%20sin%5Ctheta)
,
we can write
![S_x(t_s) = - \frac{v^2 sin2\theta}{g}](https://tex.z-dn.net/?f=S_x%28t_s%29%20%3D%20-%20%5Cfrac%7Bv%5E2%20sin2%5Ctheta%7D%7Bg%7D)
Since g is negative, the maximum of this function occurs for
![sin2\theta=1](https://tex.z-dn.net/?f=sin2%5Ctheta%3D1)
, and this happens when
![\theta=45^{\circ}](https://tex.z-dn.net/?f=%5Ctheta%3D45%5E%7B%5Ccirc%7D)
.