The surface zone :)))))))))
Answer:
What will be the increase in frequency if these waves are reflected = 2979.63Hz
Explanation:
The detailed steps and appropriate formula is as shown in the attached file.
<span> <span>We will need to work with the components of the velocity, in the x and the y direction. We will say up is positive so g is -9.81 m/s^2.
Given that the angle was 32 degrees:
Velocity up (in the y direction) is 55 m/s * sin 32 = 29.15 m/s
And
Velocity forward (in the x direction) is 55 m/s * cos 32 = 46.64 m/s
The acceleration of gravity, -9.81 m/s2 continuously decreases the velocity in the y direction. At the maximum height, the velocity will be zero. This should make sense, for as soon as the decreasing velocity becomes negative, the arrow will start to fall.
We have v = v(0) + at
And we set this to zero and solve for t:
0 = 29.15 + -9.81t
9.81t = 29.15
t = 2.97 seconds
To calculate height at this point, we use the equation that calculates position based on time, acceleration, and initial velocity (we could use an alternate too, an equation derived from the one we are now using and v = v(0) + at.
x = x(0) + v(0)t + (1/2)at^2
x = 0 + 29.15 * 2.97 + 0.5 9.81 (2.97)^2
x = 43.30 m
For a projectile, the plot of distance traveled in the upward direction is a parabola, and it takes the same amount of time to come down as it did to go up.
We can double 2.97 to get the time of impact on the target at 2(2.97) = 5.94 seconds
(Alternately, if you like, you can solve
0 = 0 + 29.15t + 0.5 9.81 t^2
And find that the two roots are 0 and 5.94).
http://www.math.com/students/calculators... will do the quadratic for you.
Given a horizontal velocity of 46.64 m/s, we can calculate
46.64 m/s (5.94 s) = 277 m for the distance of the target.</span></span>
