Question

A rescue plane wants to drop supplies to isolated mountain climbers on a rocky ridge 235 m below. The horizontal velocity of the plane is 250 km/h (69.4 m/s). Rescue plane releases the supplies a horizontal distance of 425 m in advance of the mountain climbers.

Part A: What vertical velocity (up or down) should the supplies be given so that they arrive precisely at the climbers' position (see the figure)?

Part B: With what speed do the supplies land?

Answer 1

In this solution are

Given data ....

a)

V_x = horizontal velocity = 69.4 m/s

t = time taken

X = horizontal distance travelled = 425 m

t = X/V_x = 425 / 69.4 = 6.12 sec

consider the vertical motion :

V_y = initial vertical velocity

a = acceleration = 9.8 m/s²

Y = vertical displacement = 235 m

t = time taken = 6.12 sec

using the equation

Y = V_y t + (0.5) a t²

235 = V_y (6.12) + (0.5) (9.8) (6.12)²

V_y = 8.41 m/s

velocity is in down direction

B)

V_fy = V_y + at = 8.41 + 9.8 x 6.12 = 68.4 m/s

V_fx = V_x = 69.4 m/s

V = speed = sqrt(Vfx² + Vfy²)

= sqrt((68.4)² + (69.4)²)

= 97.44 m/s