Respuesta :
1) The maximum distance can be reached when the dart is shot with an angle of [tex]\theta=45^{\circ}[/tex] 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 [tex]v_i = 5.84~m/s[/tex] and it is an uniformly accelerated motion on the y-axis (vertical direction), with the gravitational acceleration [tex]g=9.81~m/s^2[/tex] acting downwards. So we can write the laws of motion on both directions:
[tex]S_x(t) =( v_i \cos{\theta}) t[/tex]
[tex]S_y(t) = (v_i \sin{\theta})t - \frac{1}{2} gt^2[/tex]
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 [tex]S_y(t)=0[/tex]:
[tex] (v_i \sin{\theta})t - \frac{1}{2} gt^2=0[/tex]
From this we find two solutions: [tex]t=0~s[/tex], corresponding to the beginning of the motion (so we are not interested in this one), and
[tex]t= \frac{2 v_i \sin{\theta}}{g} =0.84~s[/tex]
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 [tex]t=0.84~s[/tex] inside the equation of [tex]S_x(t)[/tex] written at point 2:
[tex]S_x(0.84~s)= 5.84~m/s \cdot \cos{45^{\circ}} \cdot 0.84~s }=3.47~m[/tex]
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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:
[tex]S_x(t) = v cos{\theta} t[/tex]
[tex]S_y(t) = v sin{\theta} t + \frac{1}{2} g t^2[/tex]
where t is the time, [tex]\theta[/tex] is the initial angle of the dart, and [tex]g= -9.81 m/s^2[/tex] is the gravitational acceleration.
The maximum horizontal distance can be found by requiring that [tex]S_y=0[/tex]. This condition occurs twice: when the motion starts ([tex]t=0[/tex]) and when the dart falls to the ground (let’s call [tex]t_s[/tex] the time at which this happens). Therefore we can find [tex]t_s[/tex] by requiring [tex]S_y(t_s)=0[/tex], i.e.:
[tex]v sin{\theta} t_s + \frac{1}{2} g t_s^2 =0[/tex]
which has two solutions: [tex]t_s=0[/tex] (beginning of the motion) and [tex]t_s = - \frac{2 v sin{\theta}}{g}[/tex].So we can find the maximum horizontal distance covered by the dart by substituting this [tex]t_s[/tex] into the law of motion for [tex]S_x(t)[/tex]:
[tex]S_x(t_s) = v cos\theta t_s = -\frac{2 v^2 cos\theta sin\theta}{g}[/tex]
Since
[tex]sin2\theta = 2 cos\theta sin\theta[/tex],
we can write
[tex]S_x(t_s) = - \frac{v^2 sin2\theta}{g}[/tex]
Since g is negative, the maximum of this function occurs for [tex]sin2\theta=1[/tex], and this happens when [tex]\theta=45^{\circ}[/tex].
2) The motion of the dart is an uniform motion on the x-axis (horizontal direction) with constant velocity [tex]v_i = 5.84~m/s[/tex] and it is an uniformly accelerated motion on the y-axis (vertical direction), with the gravitational acceleration [tex]g=9.81~m/s^2[/tex] acting downwards. So we can write the laws of motion on both directions:
[tex]S_x(t) =( v_i \cos{\theta}) t[/tex]
[tex]S_y(t) = (v_i \sin{\theta})t - \frac{1}{2} gt^2[/tex]
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 [tex]S_y(t)=0[/tex]:
[tex] (v_i \sin{\theta})t - \frac{1}{2} gt^2=0[/tex]
From this we find two solutions: [tex]t=0~s[/tex], corresponding to the beginning of the motion (so we are not interested in this one), and
[tex]t= \frac{2 v_i \sin{\theta}}{g} =0.84~s[/tex]
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 [tex]t=0.84~s[/tex] inside the equation of [tex]S_x(t)[/tex] written at point 2:
[tex]S_x(0.84~s)= 5.84~m/s \cdot \cos{45^{\circ}} \cdot 0.84~s }=3.47~m[/tex]
--------------------------
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:
[tex]S_x(t) = v cos{\theta} t[/tex]
[tex]S_y(t) = v sin{\theta} t + \frac{1}{2} g t^2[/tex]
where t is the time, [tex]\theta[/tex] is the initial angle of the dart, and [tex]g= -9.81 m/s^2[/tex] is the gravitational acceleration.
The maximum horizontal distance can be found by requiring that [tex]S_y=0[/tex]. This condition occurs twice: when the motion starts ([tex]t=0[/tex]) and when the dart falls to the ground (let’s call [tex]t_s[/tex] the time at which this happens). Therefore we can find [tex]t_s[/tex] by requiring [tex]S_y(t_s)=0[/tex], i.e.:
[tex]v sin{\theta} t_s + \frac{1}{2} g t_s^2 =0[/tex]
which has two solutions: [tex]t_s=0[/tex] (beginning of the motion) and [tex]t_s = - \frac{2 v sin{\theta}}{g}[/tex].So we can find the maximum horizontal distance covered by the dart by substituting this [tex]t_s[/tex] into the law of motion for [tex]S_x(t)[/tex]:
[tex]S_x(t_s) = v cos\theta t_s = -\frac{2 v^2 cos\theta sin\theta}{g}[/tex]
Since
[tex]sin2\theta = 2 cos\theta sin\theta[/tex],
we can write
[tex]S_x(t_s) = - \frac{v^2 sin2\theta}{g}[/tex]
Since g is negative, the maximum of this function occurs for [tex]sin2\theta=1[/tex], and this happens when [tex]\theta=45^{\circ}[/tex].
