Answer:s=0.68 m
Explanation:
Given
Inclination
Speed of block(u)=1.6 m/s
Coefficient of kinetic Friction
deceleration provided by friction=g\sin \theta -\mu _kg\cos \theta [/tex]
Using
Final velocity v=0
s=0.68 m
Two beams of coherent light travel different paths, arriving at point P. If the maximum destructive interference is to occur at
1 answer:
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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
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:
(1)
where
s = -1.80 m is the vertical displacement of the shot to reach the ground (negative = downward)
is the initial vertical velocity, where
u = 12.0 m/s is the initial speed
is the angle of projection
So
is the acceleration due to gravity (downward)
Substituting the numbers, we get
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
where
is the horizontal velocity
t = 1.778 s is the time of flight
Solving, we find
B)
In this second case,
So the vertical velocity is
So the equation for the vertical motion becomes
Solving for t, we find that the time of flight is
t = 1.851 s
The horizontal velocity is
So, the range of the shot is
C)
In this third case,
So the vertical velocity is
So the equation for the vertical motion becomes
Solving for t, we find that the time of flight is
t = 1.925 s
The horizontal velocity is
So, the range of the shot is
m
D)
In this 4th case,
So the vertical velocity is
So the equation for the vertical motion becomes
Solving for t, we find that the time of flight is
t = 1.981 s
The horizontal velocity is
So, the range of the shot is
E)
From the previous parts, we see that the maximum range is obtained when the angle of releases is .
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
The solutions of this quadratic equation are
This is the time of flight: so, the horizontal range is
It can be found that the maximum of this function is obtained when the angle is
Therefore in this problem, the angle which leads to the maximum range is
Learn more about projectile motion:
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