Question

A relief airplane is delivering a food package to a group of people stranded on a very small island. The island is too small for the plane to land on, and the only way to deliver the package is by dropping it. The airplane flies horizontally with constant speed of 442km/hour at an altitude of 575m . For all parts, assume that the "island" refers to the point at a distance D from the point at which the package is released. Ignore the height of this point above sea level. Assume that the acceleration due to gravity is g = 9.80m/s^2 .
A) After a package is ejected from the plane, how long will it take for it to reach sea level from the time it is ejected? Assume that the package, like the plane, has an initial velocity of 442 km/hour in the horizontal direction.
Express your answer numerically in seconds. Neglect air resistance.
B) If the package is to land right on the island, at what horizontal distance D from the plane to the island should the package be released?
Express the distance numerically in meters.
C) What is the speed vf of the package when it hits the ground?
Express your answer numerically in meters per second.

Answer 1

speed of the plane is given as

$v = 442 km/h = 122.8 m/s$

height of the plane is

$h = 575 m$

acceleration due to gravity is

$g = 9.80 m/s^2$

part a)

when package is dropped its vertical speed is zero so we can use kinematics to find the time of drop

$\delta y = v_y * t + \frac{1}{2}gt^2$

$575 = 0 + \frac{1}{2}*9.80*t^2$

$t = 10.83 s$

Part b)

Horizontal distance moved by the package is given by

$\delta x = v_x * t$

$\delta x = 122.8 * 10.83$

$\delta x = 1330.3 m$

Part c)

final speed in x direction

$v_x = 122.8 m/s$

final speed in y direction

$v_y = 9.8*10.83 = 106.13 m/s$

so net speed as it hit the ground will be

$v = \sqrt{v_y^2 + v_x^2}$

$v = \sqrt{122.8^2 + 106.13^2}$

$v = 162.3 m/s$