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. Assume the plane is traveling horizontally with a speed of 245 km/h (68.1 m/s).

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 (Figure 3-37b)? (Assume up is positive.)?

With what speed do the supplies land in the latter case?

Explanation / Answer

x direction

x = v0x t

t = 425/68.1=6.24 s

y direction

y = y0 + v0y t + 1/2 a t^2

0 = 235 + v*6.24 - 0.5*9.81*6.24^2

v=-7.05 m/s

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