# 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,...

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?

What horizontal distance is traveled by this
package?

krishna-agrawala | College Teacher | (Level 3) Valedictorian

Posted on

The first part of the answer regarding first package requires no calculation. The package will have the same horizontal speed as the plane, and therefore, when the package hits the ground, the plane will be directly above the package.

To calculate the horizontal distance of the package from the drop point, we first calculate t, the time taken for the drop using following equation:

t = (2s/a)^1/2

Where s = vertical distance travelled during drop = 2770 m

and a = acceleration = 9.8 m/s^2

Therefore: t = (2*2770/9.8)^1/2 = 23.7762 s (approximately)

Horizontal distance of package

= t*horizontal speed = 23.7762*315 = 7489.503 m (approximately)

Now taking up the solution for the second package:

(Please note that same symbols have been used for the solution to problem relating to the first and second package, but their values are different.)

It is assumed that the package is thrown at vertical speed of 52 m/s with respect to the plane. Therefore the package will have a constant horizontal component of speed equal to speed of plane. This is given as 315 m/s.

Given: Initial vertical velocity of package = u = 52 m/s

vertical distance travelled = s = 2770

vertical acceleration = a = 9.8 m/s^2

Final vertical velocity = v = ?

Time taken for the package to hit the ground = t = ?

To find v we use the expression: v^2 = u^2 + 2as

Therefore v = (u^2 + 2as)^1/2

= (52^2 + 2*9.8*2770)^1/2 = 238.7384 (approximately)

Also: t = (v - u)/a = (238.7384 - 52)/9.8 = 19.0549 (approximately)

Horizontal distance travelled by package = t*(horizontal speed)

= 19.0549*315 = 6002.3041 m (approximately)

Magnitude of velocity of package at the time of drop as seen by an observer from ground is given by the following equation.

Magnitude of net velocity = [(horizontal velocity)^2 = vertical velocity)^2]^1/2

= (315^2 +52^2)^1/2 = 319.2632 m/s approximately

neela | High School Teacher | (Level 3) Valedictorian

Posted on

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.