Respuesta :
Given,
The initial velocity of the plane, u=175 m/s
The angle of projection of the plane, θ=35.5°
The time interval, t=48.0 s
The x-component of the initial velocity is,
[tex]u_x=u\cos \theta[/tex]On substituting the known values,
[tex]u_x=175\times\cos 35.5^0=142.5\text{ m/s}[/tex]The y-component of the initial velocity is
[tex]u_y=u\sin \theta_{}[/tex]On substituting the known values,
[tex]u_y=175\times sin35.5^0=101.6\text{ m/s}[/tex]The x-component of the final velocity will remain the same as there is no acceleration on the plane in the x-direction.
The y-component of the final velocity can be calculated as,
[tex]v_y=u_y+gt[/tex]Here, the acceleration due to gravity will be a negative value, as the gravity will be acting downwards. On substituting the known values,
[tex]\begin{gathered} v_y=101.6-9.8\times48.0_{} \\ =-368.8\text{ m/s} \end{gathered}[/tex]Therefore the magnitude of the final velocity is given by,
[tex]v=\sqrt[]{u^2_x+v^2_y}[/tex]On substituting the known values,
[tex]\begin{gathered} v=\sqrt[]{142.5^2+(-368.8)^2} \\ =395\text{ m/s} \end{gathered}[/tex]Therefore the final velocity of the plane will be 395 m/s
As the sun is directly above the plane, the distance traveled by the shadow of the plane will be equal to the range of the projectile motion of the plane.
The range is given by,
[tex]R=\frac{u^2\sin 2\theta}{g}[/tex]On substituting the known values,
[tex]\begin{gathered} R=\frac{175^2\times\sin(2\times35.5)}{9.8} \\ =2954.75\text{ m} \end{gathered}[/tex]Therefore the total distance traveled by the shadow is 2954.75 m
