A plane is ﬂying horizontally with speed 232 m/s at a height 5240 m above the ground, when a package is droppe?The acceleration of gravity is 9.8 m/s2. Neglecting air resistance, when the package hits the ground, the plane will be1. ahead of the package.2. behind the package.3. directly above the package.What is the horizontal distance from the release point to the impact point?Answer in units of mA second package is thrown downward from the plane with a vertical speed v1 = 76 m/s.What is the magnitude of the total velocity of the package at the moment it is thrown as seen by an observer on the ground?Answer in units of m/sWhat horizontal distance is traveled by this package?Answer in units of m

a.The first part of the question regarding first package is straightforward. The package has the same horizontal speed as the plane, and therefore, when the package hits the ground, the plane will be above the package dropped.

b.The height of drop = 5240 m

The velocity of the plane=232 m/s

[tex]s=\frac{1}{2} *gt^2[/tex]

The packages has horizontal component of velocity =232 m/s and vertical component of velocity = gt at a time t.

The time t to hit the ground is given by:

[tex]s=\frac{1}{2} *gt^2[/tex] = vertical displacement s = 5240 m Solving for t we get:

t=(5240*2/9.8)^(1/2)=32.701secs. This is the time for the package to reach the ground.

Therefore, the horizontal displacement travelled by the package =horizontal vel* time=32.701*232=7586.632m

Therefore, the displacement from the starting point of the drop till the package hit the ground=(horizontal displacement^2+vertical displacement^2)^(1/2)=(7586.632m^2+ 5240^2)(1/2)=9219m

The plane moves in the x direction by this time = 32.701*232=7586.632m which is equal to the horizontal displacement,neglecting air resistance. So, the package and plane are at the same horizontal displacement ( neglecting air resistance ).

Second package:

Initial vertical velocity component of the package v1 = 76 m/s

The horizontal component of velocity of the package = 232m/s

Therefore the initial magnitude of the velocity = (horizontal vel^2+vertical vel^2)^(1/2)= (232^2+76^2)(^1/2)=244.13m/s

its in a direction declined to horizontal by an angle

tan^-1( 76/232)

= tangent inverse( 76/232) =18.13 deg

b.244.13m/s in the direction 18.13 deg

The vertical displacement of the package = 5240 m

Time to get to the ground =ut+(1/2)gt^2, where u is the initial vertical velocity towards earth , t is the time to reach the ground and g is the acceleration due to gravity. Therefore, 5240 =76t+0.5(9.8)t^2

t= 25.85s

=The horizontal displacement during this time = 232* 25.85s=5997.2m

The total displacement = sqrt( horizontal displacement^2+ vertical displacement ^2)

sqrt(5997.2^2+5240^2 )=sqrt(27457600m)

The total displacement= 7963.91m