An airplane with a speed of 97.5 m/s is climbing upward at an angle of
50.0∘
with respect to the horizontal. When the plane's altitude is 732 m, the pilot releases a package. (a) Calculate the distance along the ground, measured from a point directly beneath the point of release, to where the package hits the earth. (b) Relative to the ground, determine the angle of the velocity vector of the package just before impact.

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Respuesta :

aristocles aristocles · Answer 1 · 2019-08-04T11:57:47+02:00

Answer:

Part A)

d = 1378.8 m

Part b)

[tex]\theta = 66 degree[/tex]

Explanation:

The two components of the velocity of the plane is given as

[tex]v_y = 97.5 sin50 = 74.7 m/s[/tex]

[tex]v_x = 97.5 cos50 = 62.7 m/s[/tex]

now the height of the packet when it is released is given as

[tex]H = 732 m[/tex]

now the time taken by it to reach the ground is given as

[tex]\delta y = v_y t + \frac{1}{2}a_y t^2[/tex]

now we have

[tex]-732 = 74.7 t - \frac{1}{2}(9.8) t^2[/tex]

after solving above equation

t = 22 s

Part a)

Now the distance covered by the package in the same time when it reaches to the ground

[tex]d = v_x t[/tex]

[tex]d = (62.7 m/s)(22 s) = 1378.8 m[/tex]

Part b)

At the time it will reach the ground the final velocities are given as

[tex]v_{fy} = v_y + at = (74.7) + (-9.81)(22) = -141.1 m/s[/tex]

[tex]v_{fx} = v_x = 62.7 m/s[/tex]

now the angle at which it will strike the ground is given as

[tex]tan\theta = \frac{-141.1}{62.7} = -2.25[/tex]

[tex]\theta = 66 degree[/tex]