A plane is flying horizontally with speed 315ms at a height

A plane is flying horizontally with speed 315m/s at a height 2770m above the ground, when a package is dropped from the plane.

The acceleration of gravity is 9.8m/s^2. Neglecting air resistance, when the package hits the ground, the plane will be.

1) behind the package. 2) ahead of the package. 3) directly above the package.

What is the horizontal distance from the release point to the impact point? Answer in units of m.

A second package is thrown downward from
the plane with a vertical speed v1 = 52 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/s.

What horizontal distance is traveled by this
package?
Answer in units of m.

Solution

The height of drop = 2770m

The velocity of the plane=315m/s

The packages has horizontal component of velocity =315m/s and vertical component of velocity = gt at at time t.

The time t to hit the ground is given by: (1/2)gt^2= vertical displacement s = 2770m. Solving for t we get:

t=(2770*2/9.8)^(1/2)=23.7762secs. This is the time for the package to reach the ground.

Therefore, the horizontal displacement travelled by the package =horizontal vel* time=315*23.7762=7489.4926m.

Therefore, the displacement from the starting point of the drop till the package reached the ground=(horizontal displacement^2+vertical displacement^2)^(1/2)=(7489.4962^2+2770^2)(1/2)=7985.3240m

The plane moves (horizontally) by this time = 315*23.7762=7489.4926m which is equal to the horizotal diplacement,neglecting air resistance. So, theoretically the package and plane are at the same horizontal diplacement ( neglecting air resistance ).

Second package:

Initial vertical velocity component of the package = 52m/s

The horizontal component of velocity of the package = 315m/s

Therefore the initial magnitude of the velocity = (horizontal vel^2+vertical vel^2)^(1/2)= (315^2+52^2)(^1/2)=319.2632m/s and in a direction declined to horizontal by an angle tangent inverse(vertical velocity /horizontal velocity) = tangent inverse(52/315) =9.3738 degree.

The vertical displacement of the package = 2770m

Time to reach the ground is given by 2770 =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, 2770 =52t+0.5(9.8)t^2 or 4.9t^2+52t-2770=0

t= (-52 + or - sqrt(52^2-4*4.9*(-2770))/(2(4.9))

=19.0549s. The horizontal displacement during this time = 315*19.0549=6002.3041m

The total displacement from the starting point till the package reached the ground= sqrt( hor displacement^2+ vertical displacement ^2)= sqrt(6002.3041^2+2770^2)=6610.6395 m

Hope this helps.