Question

An airplane with a speed of 88.0 m/s is climbing upward at an angle of 40.8 ° with respect to the horizontal. When the plane's altitude is 515 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

given
vo = 88 m/s
theta = 40.8 degrees
h = 515 m
vox = 88*cos(40.8) = 66.6 m/s
voy = 88*sin(40.8) = 57.5 m/s

a) let t is the time taken for the package to reach the ground.

use,

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

-515 = 57.5*t - (1/2)*9.8*t^2

on solving the above equation we get, t = 17.68 s

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

x = vox*t

= 66.6*17.68

= 1777 m

b) just before touching the ground, vx = vox

= 66.6 m/s

vy = voy - g*t

= 57.5 - 9.8*17.68

= -116 m/s

angle made by velocity vector just before impact, theta = tan^-1(vy/vx)

= tan^-1(116/66.6)

= 60.1 degrees