Question

A plane is flying horizontally at 121 m/s relative to the air, and the air is moving relative to the ground horizontally at 49.0m/s. (a)At what angle relative to the wind must the plane fly so that when it drops a package from a height of 5560 m the package travels equal horizontal and vertical distances before hitting the ground? Assume no significant drag on the package and that the package is released from rest relative to the plane.(b)If the plane is instead flying at an angle of 131°relative to the wind, then at what angle will the projectile hit the ground 5560m below? Assume no significant drag on the package and that the package is released from rest relative to the plane. (c)If the plane is instead flying at an angle of 12.0°relative to the wind, then what is the distance between the package and the plane at the moment the package hits the ground 5560m below?

Answer 1

velocity of the plane = 121 m/s relative to air

air velocity = 49 m/s relative ground

height of the drop = 5560 m

no air drag

The plane is moving horizontal.

initial vertical velocity =0

time of fall t = sqrt(2h/g) = sqrt( 2*5560/9.8) = 33.69 s

horizontal displacement of the package = 5560 m in 33.69 s

horizontal velocity = 5560/33.69 = 165.05 m/s

The package must have a resultant horizontal velocity of 165.05 relative to ground.

if the angle between the plane direction and wind direction =

resultant velocity =sqrt(121² + 49² + 2*121*49*Cos() ) = 165.05

= 30.82 , the plane make an angle so that the package travels equal horizontal distance as the height

b) angle = 131deg.

resultant velocity of the package when dropped ( horizontal)

= sqrt(121² + 49² + 2*121*49*Cos(131) ) = 96.24 m/s

vertical velocity of the package when hits the ground = sqrt ( 2gh) = sqrt( 2*9.8*5560) = 330.12 m/s

horizontal velocity = 96.24 m/s

hitting angle with the ground :

Tan (p) = u_x/u_y = 96.24/330.12

p = 16.26 , the package makes an angle 16.26^o with the ground

c) angle = 12^o  

resultant velocity (horizontal) = sqrt(121² + 49² + 2*121*49*Cos(12) ) = 169.24 m/s.

There is no air drag on the package. Its horizontal velocity remains same as the plane. Horizontal distance between the plane and the package is 0.

vertical distance = 5560m.

distance between the package and the plane = 5560 m