Two planes each dropping an empty fuel tank. At the moment of release each plane

Question

Two planes each dropping an empty fuel tank. At the moment of release each plane has the same speed of 158 m/s, and each tank is at the same height of 3.61 km above the ground. Although the speeds are the same, the velocities are different at the instant of release, because one plane is flying at an angle of 15.0° above the horizontal and the other is flying at an angle of 15.0° below the horizontal. Find the (a) magnitude and (b) direction of the velocity with which the fuel tank hits the ground if it is from plane A. Find the (c) magnitude and (d) direction of the velocity with which the fuel tank hits the ground if it is from plane B. In each part, give the direction as a positive angle with respect to the horizontal.

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Explanation / Answer

plane A:

velocity of the fuel tank at relase=velocity of plane A=158 m/s , at 15 degree above horizontal

then initial vertical velocity=158*sin(15)=40.89 m/s

vertical acceleration=-9.8 m/s^2 (downwards)

vertical distance to be covered=-3610 m (downward, hence taken as negative)

let time taken be t.

then 40.89*t-0.5*9.8*t^2=-3610
t=31.6341 seconds

then vertical velocity of the plane when it hits ground=40.89-9.8*t=-269.12 m/s

horizontal velocity will remain the same as initial as there is no force in horizontal direction

hence horizontal velocity=158*cos(15)=152.61 m/s

hence net speed=sqrt(152.61^2+269.12^2)=309.38 m/s

angle with horizontal=arctan(-269.12/152.61)=-60.443 degrees

i.e. the angle is 60.443 degree below horizontal.

for plane B:

velocity of the fuel tank at relase=velocity of plane A=158 m/s , at 15 degree below horizontal

then initial vertical velocity=158*sin(-15)=-40.89 m/s

vertical acceleration=-9.8 m/s^2 (downwards)

vertical distance to be covered=-3610 m (downward, hence taken as negative)

let time taken be t.

then -40.89*t-0.5*9.8*t^2=-3610
t=23.2892 seconds

then vertical velocity of the plane when it hits ground=-40.89-9.8*t=-269.12 m/s

horizontal velocity will remain the same as initial as there is no force in horizontal direction

hence horizontal velocity=158*cos(15)=152.61 m/s

hence net speed=sqrt(152.61^2+269.12^2)=309.38 m/s

angle with horizontal=arctan(-269.12/152.61)=-60.443 degrees

i.e. the angle is 60.443 degree below horizontal.

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