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...

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