posted by Me
A passenger in a helicopter traveling upwards at 19 m/s accidentally drops a package out the window. If it takes 14 seconds to reach the ground, how high to the nearest meter was the helicopter when the package was dropped?
To the nearest meter what was the maximum height of the package above the ground in the previous problem?
Height change after release
= 19 t - (g/2)t^2
= -H when t = 14 s.
H is the height above ground where it is released and g is the acceleration of gravity.
Solve for H.
266 - 960 = -H
H = 694 m
It rises a distance h above H before coming back down.
g h = V^2/2 where V is the initial upward speed. (From energy considerations)
h = 18.4 m
max height = 694 + 18 = 712 m