Respuesta :
A) Horizontal range: 16.34 m
B) Horizontal range: 16.38 m
C) Horizontal range: 16.34 m
D) Horizontal range: 16.07 m
E) The angle that gives the maximum range is [tex]41.9^{\circ}[/tex]
Explanation:
A)
The motion of the shot is a projectile motion, so we can analyze separately its vertical motion and its horizontal motion.
The vertical motion is a uniformly accelerated motion, so we can use the following suvat equation to find the time of flight:
[tex]s=u_y t + \frac{1}{2}at^2[/tex] (1)
where
s = -1.80 m is the vertical displacement of the shot to reach the ground (negative = downward)
[tex]u_y = u sin \theta[/tex] is the initial vertical velocity, where
u = 12.0 m/s is the initial speed
[tex]\theta=40.0^{\circ}[/tex] is the angle of projection
So
[tex]u_y=(12.0)(sin 40.0^{\circ})=7.7 m/s[/tex]
[tex]a=g=-9.8 m/s^2[/tex] is the acceleration due to gravity (downward)
Substituting the numbers, we get
[tex]-1.80 = 7.7t -4.9t^2\\4.9t^2-7.7t-1.80=0[/tex]
which has two solutions:
t = -0.21 s (negative, we ignore it)
t = 1.778 s (this is the time of flight)
The horizontal motion is instead uniform, so the horizontal range is given by
[tex]d=u_x t[/tex]
where
[tex]u_x = u cos \theta=(12.0)(cos 40^{\circ})=9.19 m/s[/tex] is the horizontal velocity
t = 1.778 s is the time of flight
Solving, we find
[tex]d=(9.19)(1.778)=16.34 m[/tex]
B)
In this second case,
[tex]\theta=42.5^{\circ}[/tex]
So the vertical velocity is
[tex]u_y = u sin \theta = (12.0)(sin 42.5^{\circ})=8.1 m/s[/tex]
So the equation for the vertical motion becomes
[tex]4.9t^2-8.1t-1.80=0[/tex]
Solving for t, we find that the time of flight is
t = 1.851 s
The horizontal velocity is
[tex]u_x = u cos \theta = (12.0)(cos 42.5^{\circ})=8.85 m/s[/tex]
So, the range of the shot is
[tex]d=u_x t = (8.85)(1.851)=16.38 m[/tex]
C)
In this third case,
[tex]\theta=45^{\circ}[/tex]
So the vertical velocity is
[tex]u_y = u sin \theta = (12.0)(sin 45^{\circ})=8.5 m/s[/tex]
So the equation for the vertical motion becomes
[tex]4.9t^2-8.5t-1.80=0[/tex]
Solving for t, we find that the time of flight is
t = 1.925 s
The horizontal velocity is
[tex]u_x = u cos \theta = (12.0)(cos 45^{\circ})=8.49 m/s[/tex]
So, the range of the shot is
[tex]d=u_x t = (8.49)(1.925)=16.34[/tex] m
D)
In this 4th case,
[tex]\theta=47.5^{\circ}[/tex]
So the vertical velocity is
[tex]u_y = u sin \theta = (12.0)(sin 47.5^{\circ})=8.8 m/s[/tex]
So the equation for the vertical motion becomes
[tex]4.9t^2-8.8t-1.80=0[/tex]
Solving for t, we find that the time of flight is
t = 1.981 s
The horizontal velocity is
[tex]u_x = u cos \theta = (12.0)(cos 47.5^{\circ})=8.11 m/s[/tex]
So, the range of the shot is
[tex]d=u_x t = (8.11)(1.981)=16.07 m[/tex]
E)
From the previous parts, we see that the maximum range is obtained when the angle of releases is [tex]\theta=42.5^{\circ}[/tex].
The actual angle of release which corresponds to the maximum range can be obtained as follows:
The equation for the vertical motion can be rewritten as
[tex]s-u sin \theta t + \frac{1}{2}gt^2=0[/tex]
The solutions of this quadratic equation are
[tex]t=\frac{u sin \theta \pm \sqrt{u^2 sin^2 \theta+2gs}}{-g}[/tex]
This is the time of flight: so, the horizontal range is
[tex]d=u_x t = u cos \theta (\frac{u sin \theta \pm \sqrt{u^2 sin^2 \theta+2gs}}{-g})=\\=\frac{u^2}{-2g}(1+\sqrt{1+\frac{2gs}{u^2 sin^2 \theta}})sin 2\theta[/tex]
It can be found that the maximum of this function is obtained when the angle is
[tex]\theta=cos^{-1}(\sqrt{\frac{2gs+u^2}{2gs+2u^2}})[/tex]
Therefore in this problem, the angle which leads to the maximum range is
[tex]\theta=cos^{-1}(\sqrt{\frac{2(-9.8)(-1.80)+(12.0)^2}{2(-9.8)(-1.80)+2(12.0)^2}})=41.9^{\circ}[/tex]
Learn more about projectile motion:
brainly.com/question/8751410
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