A helicopter is flying in a straight line over a level field at a constant speed

Question

A helicopter is flying in a straight line over a level field at a constant speed of 6.6 m/s and at a constant altitude of 9.9 m. A package is ejected horizontally from the helicopter with an initial velocity of 14.0 m/s relative to the helicopter, and in a direction opposite the helicopter's motion.


1) Find the initial velocity of the package relative to the ground.


2) What is the horizontal distance between the helicopter and the package at the instant the package strikes the ground?


3) What angle does the velocity vector of the package make with the ground at the instant before impact, as seen from the ground?

Explanation / Answer

Velocity of helicopter = Vh = 6.6m/s

Vpackage w.r.t. ground = V

Vpackage w.r.t. helicopter = Vph

we can write that V = Vh+Vph = 6.6-13.0 = -6.4 (negative sign indicates opposite direction)

Initial velocity of package w.r.t. ground = 6.4m/s

Time taken by the package to hit the ground is (2*h/g) = (2*8.7/9.81) = 1.33 sec

Horizontal distance between helicopter and package after 1.33 sec = Vph*1.33 = 17.30m

vertical velocity just before impact with ground = g*t = 9.81*1.33 = 13.05 m/s

V = -6.4i-13.05j

Angle made with x-axis = arctan(13.05/6.4) = 63.9 degree (arctan = taninverse)

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