Question

A helicopter is flying in a straight line over a level field at a constant speed of 5.70 m/s and at a constant altitude of 8.90 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, Find the initial speed of the package relative to the ground, What is the horizontal distance between the helicopter and the package at the instant the package strikes the ground? What angle does the velocity vector of the package make with the ground at the instant before impact, as seen from the ground?

Answer 1

Initial horizontal speed = vx = 5.70 m/s

Height above ground = y = 8.9 m

Initial velocity of package with respect to helicopter = xph = = -14 m

Initial velocity of package with respect to helicopter = Initial velocity of packet with respect to Earth - Speed of helicopter

-14.0 =v - 5.70 m/s

v=- 8.3 m/s

speed = 8.3 m/s

(b) time taken to strike the ground = t = ?

vertical distance travelled = y = 8.90 m

initial vertical velocity = voy = 0 m/s

acceleration = a = -9.81 m/s^2

Using

y = voyt +0.5at^2

-8.9 = 0+ 0.5(-9.81)*t^2

t = 1.81 s

in this time, package hits ground, and helicopter travels d = speed of helicopter x time = 5.70 x 1.81 = 10.3 m

package covers horizontal distance = d' = speed of package x time = 8.3 x 1.81 = 15 m

horizontal distance between package and helicopter when package hits ground = 15 - 10.3 = 4.7 m

(c) angle = arctan( final vertical velocity/ horizontal velocity)

final vertical velocity = v'

initial vertical velocity = 0

acceleration = a - -9.81 m/s^2

t = times = 1.81 s

v' -v = at

v' - 0 = -9.81 x 1.81 = -17.8 m/s

angle = arctan ( -17.8/8.3) = 65 degrees with ground