Answer:
a
[tex]D = 1162.7 \ m [/tex]
b
[tex]\beta =- 65.55^o[/tex]
Explanation:
From the question we are told that
The speed of the airplane is [tex]u = 92.3 \ m/s[/tex]
The angle is [tex]\theta = 51.1^o[/tex]
The altitude of the plane is [tex]d = 532 \ m[/tex]
Generally the y-component of the airplanes velocity is
[tex]u_y = v * sin (\theta )[/tex]
=> [tex]u_y = 92.3 * sin ( 51.1 )[/tex]
=> [tex]u_y = 71.83 \ m/s[/tex]
Generally the displacement traveled by the package in the vertical direction is
[tex]d = (u_y)t + \frac{1}{2}(-g)t^2[/tex]
=> [tex] -532 = 71.83 t + \frac{1}{2}(-9.8)t^2[/tex]
Here the negative sign for the distance show that the direction is along the negative y-axis
=> [tex]4.9t^2 - 71.83t - 532 = 0[/tex]
Solving this using quadratic formula we obtain that
[tex]t = 20.06 \ s[/tex]
Generally the x-component of the velocity is
[tex]u_x = u * cos (\theta)[/tex]
=> [tex]u_x = 92.3 * cos (51.1)[/tex]
=> [tex]u_x = 57.96 \ m/s[/tex]
Generally the distance travel in the horizontal direction is
[tex]D = u_x * t[/tex]
=> [tex]D = 57.96 * 20.06 [/tex]
=> [tex]D = 1162.7 \ m [/tex]
Generally the angle of the velocity vector relative to the ground is mathematically represented as
[tex]\beta = tan ^{-1}[\frac{v_y}{v_x } ][/tex]
Here [tex]v_y[/tex] is the final velocity of the package along the vertical axis and this is mathematically represented as
[tex]v_y = u_y - gt[/tex]
=> [tex]v_y = 71.83 - 9.8 * 20.06[/tex]
=> [tex]v_y = -130.05 \ m/s [/tex]
and v_x is the final velocity of the package which is equivalent to the initial velocity [tex]u_x[/tex]
So
[tex]\beta = tan ^{-1}[-130.05}{57.96 } ][/tex]
[tex]\beta =- 65.55^o[/tex]
The negative direction show that it is moving towards the south east direction
