Answer:
The plane will be 4795.23 meters from where he threw the package.
Explanation:
Data:
v = 167 m/s
h = 4040 m
g = 9.8m/s²
package:
after it is dropped there is no horizontal force, so...
[tex]X - Xo = Vt[/tex]
t = [tex]\frac{X - Xo}{V}[/tex] I
in the vertical we have only [tex]P = mg[/tex]
[tex]Y - Yo = Vot -\frac{gt^{2} }{2}[/tex] but Vo = 0 because the package is dropped
[tex]Y - Yo =-\frac{gt^{2} }{2}[/tex] II
replacing I in II
[tex]Y - Yo =-\frac{g(X - Xo)^{2} }{2V^{2}}[/tex]
[tex]0 - 4040 =-\frac{9.8(X - Xo)^{2} }{2*167^{2}}[/tex]
X - Xo = 4795.23
The distance from where the plane drops the package and where it hits the ground is the same as the plane flies horizontally, as there is no acceleration at x.
