Question

What is the maximum distance we can shoot a dart, provided our toy dart gun gives a maximum initial velocity of 5.84 m/s?

Answer 1

1) The maximum distance can be reached when the dart is shot with an angle of $\theta=45^{\circ}$ 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 $v_i = 5.84~m/s$ and it is an uniformly accelerated motion on the y-axis (vertical direction), with the gravitational acceleration $g=9.81~m/s^2$ acting downwards. So we can write the laws of motion on both directions:
$S_x(t) =( v_i \cos{\theta}) t$
$S_y(t) = (v_i \sin{\theta})t - \frac{1}{2} gt^2$
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 $S_y(t)=0$:
$(v_i \sin{\theta})t - \frac{1}{2} gt^2=0$
From this we find two solutions: $t=0~s$, corresponding to the beginning of the motion (so we are not interested in this one), and 
$t= \frac{2 v_i \sin{\theta}}{g} =0.84~s$

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 $t=0.84~s$ inside the equation of $S_x(t)$ written at point 2:
$S_x(0.84~s)= 5.84~m/s \cdot \cos{45^{\circ}} \cdot 0.84~s }=3.47~m$

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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:
$S_x(t) = v cos{\theta} t$
$S_y(t) = v sin{\theta} t + \frac{1}{2} g t^2$
where t is the time, $\theta$ is the initial angle of the dart, and $g= -9.81 m/s^2$ is the gravitational acceleration.
The maximum horizontal distance  can be found by requiring that $S_y=0$. This condition occurs twice: when the motion starts ($t=0$) and when the dart falls to the ground (let’s call $t_s$ the time at which this happens). Therefore we can find $t_s$ by requiring $S_y(t_s)=0$, i.e.:
$v sin{\theta} t_s + \frac{1}{2} g t_s^2 =0$
which has two solutions: $t_s=0$ (beginning of the motion) and $t_s = - \frac{2 v sin{\theta}}{g}$.So we can find the maximum horizontal distance covered by the dart by substituting this $t_s$ into the law of motion for $S_x(t)$:
$S_x(t_s) = v cos\theta t_s = -\frac{2 v^2 cos\theta sin\theta}{g}$ 
Since
$sin2\theta = 2 cos\theta sin\theta$,
we can write
$S_x(t_s) = - \frac{v^2 sin2\theta}{g}$
Since g is negative, the maximum of this function occurs for $sin2\theta=1$, and this happens when $\theta=45^{\circ}$.