A rescue plane wants to drop supplies to isolated mountain climbers on a rocky r

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?

Explanation / Answer

Ans:-

Let t be time to fall down for the supplies if not given any velocity in the Vertical direction, t is given by
235 = 0.5*9.8*t^2

t = sqrt[235/(0.5*9.8)] = 6.925 s
In 6.92 5 seconds the horizontal distance travelled by supplies would be

69.4x6.925 = 480.595 m.

So supplies should have been dropped 481 m in advance. when they are dropped 425 m in advance supplies would fall 56 m away. The required time of fall should be 425/69.4 =6.12 s. Obviously to reduce time of fall we must throw supplies with some vertical speed down wards. so that the time of fall should be 6.12s.

Let that downward velocity be V, so the equation governing V, g, s and t is
s = Vt +1/2 g t^2

substituting values, we have
235 = 6.12V + 0.5*9.8*6.12^2
V = 8.41m/s

The vertical component Vv of the supplies when they land is given by

Vv = V + 6.12*9.8

= 8.41 + 59.98 = 68.39m/s

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