Question

A helicopter drops a package of medicine to a remote village. The helicopter flies horizontally with a speed of 33.6 m/s. The package hits the ground at a speed of 43.5 m/s. What is the velocity of the package just before impact? (a)vx= (b)vy= (c) How high above the ground is the helicopter when it releases the package? (d) How far does the package travel horizontally as it falls to the ground?

Answer 1

Taking the downward direction as negative.

Gravitational acceleration = g = -9.81 m/s²

Speed of the helicopter = V = 33.6 m/s (Flying horizontally)

The package will have the same velocity as the helicopter at the time of release.

Initial velocity of the package = V₀ = 33.6 m/s

Initial horizontal velocity of the package = V_x0 = 33.6 m/s

Initial vertical velocity of the package = V_y0 = 0 m/s

Velocity of the package when it hits the ground = V₁ = 43.5 m/s

Horizontal velocity of the package when it hits the ground = V_x1

Vertical velocity of the package when it hits the ground = V_y1

There is no force acting on the package in the horizontal direction hence the velocity of the package in the horizontal direction doesn't change.

V_x1 = V_x0 = 33.6 m/s

V₁² = V_x1² + V_y1²

(43.5)² = (33.6)² + V_y1²

V_y1 = -27.627 m/s

Negative as it is directed downwards.

Height of the helicopter above the ground when it releases the package = H

V_y1² = V_y0² + 2gH

(-27.627)² = (0)² + (2)(-9.81)H

H = 38.9 m

Time taken by the package to hit the ground = T

H = V_y0T + gT²/2

38.9 = (0)T + (9.81)T²/2

T² = 7.93

T = 2.816 sec

Horizontal distance covered by the package = R

R = V_x0T

R = (33.6)(2.816)

R = 94.62 m

Velocity of the package before the impact = 43.5 m/s

a) Horizontal velocity of the package before the impact = 33.6 m/s

b) Vertical velocity of the package before the impact = -27.627 m/s

c) Height of the helicopter above the ground when the package was released = 38.9 m

d) Horizontal distance covered by the package = 94.62 m