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.

Paert A

If the plane is traveling horizontally with a speed of 263 km/h (73.1 m/s ), how far in advance of the recipients (horizontal distance) must the goods be dropped.

Part B

Suppose, instead, that the plane releases the supplies a horizontal distance of 425 m in advance of the mountain climbers. What vertical velocity (up or down) should the supplies be given so that they arrive precisely at the climbers' position .

Explanation / Answer

a) Time to fall t = sqrt(2h / g)

= sqrt(2 * 235m / 9.8m/s^2)

= 6.93 s

lead distance d = v*t

= 73.1 m/s * 6.93 s

= 507 m answer

b) Now the travel time is t = d / v

= 425 m / 73.1 m/s = 5.81 s

and we need a vertical velocity V such that

s = So + V*t + 0.5 at^2

0 = 235m + V*5.81 s - 0.5* 9.8m/s^2 * (5.81 s)^2

V = (0.5* 9.8m/s^2 * (5.81 s)^2 - 235 m)/5.81 s

= -11.98 m/s (that is , down) answer

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