1) A mail bag is to be dropped into a post office from an aeroplane flying horizontally with a velocity of 270km/hr at a height of 176.4m above the ground.how far must the aeroplane be from the post office at the time of dropping the bag so that it directly falls into the post office?

The first step is to convert the velocity from km/h to m/s.

1 km = 1000 m, 1 h = 3600 s, 1 km/h = 1000/3600 = 5/18 m/s

v = 270 * 5/18 = 75 m/s

When the package is dropped, it has a horizontal velocity of 75 m/s. To determine the horizontal distance that the package will move as it falls to the ground, use the following equation.

d = vi * t + ½ * a * t^2, vi is the initial vertical velocity. This is 0 m/s. a is 9.8 m/s^2.

176.4 = ½ * 9.8 * t^2

t = √(176.4/4.9) = 6 seconds

To determine the horizontal distance, use the following equation.

d = v * t = 75 * 6 = 450 meters

For the package to fall directly into the post office, it must be dropped when the plane is a horizontal distance of 450 meters from it. To determine the actual distance, use the following equation.

d = √(dh^2+ dv^2)= √(450^2 + 176.4^2) = √233,616.96

This is approximately 483 meters

hope u unerstood