Question

An airplane with a speed of 81.1 m/s is climbing upward at an angle of 45.2 ° with respect to the horizontal. When the plane's altitude is 535 m, the pilot releases a package. (a) Calculate the distance along the ground, measured from a point directly beneath the point of release, to where the package hits the earth. (b) Relative to the ground, determine the angle of the velocity vector of the package just before impact.

Answer 1

a) when the package is released from the Airplane,

vo = 81.1 m/s
theta = 45.2 m/s
h = 535 m

vox = 81.1*cos(45.2) = 57.1 m/s
voy = 81.1*sin(45.2) = 57.5 m/s

let t is the time taken for the packet to reach the ground.

use, -h = voy*t - (1/2)*g*t^2

-535 = 57.5*t - (1/2)*9.8*t^2

on solving the above equation we get

t = 17.85 s

the distance along the ground, measured from a point directly beneath the point of release, to where the package hits the earth,

x = vox*t

= 57.1*17.85

= 1019 m <<<<<<<<<<------------------Answer

b) when package hits the ground,
vx = vox

= 57.1 m/s

vy = voy - g*t

= 57.5 - 9.8*17.85

= -117 m/s

angle made by v with horizontal, theta = tan^-1(vy/vx)

= tan^-1(117/57.1)

= 64.0 degrees <<<<<<<<<<<<---------------------------Answer