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 235m below. If the plane is traveling horizontally with a speed of 250km/h (69.4 m/s), (a) how far in advance of the recipients (horizontal distance) must the goods be dropped? (b) Suppose, instead, that the plane releases the supplies a horizontal distance of 425m 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. (c) With that speed do the supplies land in the latter case?

Explanation / Answer

   for simplicity let us choose the origin tobe the location on the ground directly below the air plane at thetime the supplies    are dropped, upward is positive directionthen for the supplies we get    yo = 235 m    vyo = 0    ay = - g    y = 0 m    vx = 69.4 m / s (a)    to get the time taken for supplies toreach the ground we u8se the equation of motion    y = yo +  vyo t + (1 / 2) ay t2    0 = yo + 0 + (1 / 2) a t2    t = (- 2 yo / a)      = ....... s    the horizontal distance will be    x = vx t         = .........m (b)    so now the supplies have to travel a horizontaldistance of only 425 m    so the time of flight will be    t = 425 m / 69.4 m / s      = ......... s    again using the equaytion of motion we getthe velocity as    vyo = (y -  yo - (1 / 2) ay t2) / t          = ........ m/ s (c)    the vertical speed will be    vy = vyo +ay t         = ......... m /s    so the magnitude of the speed will be    v = (vx2 +vy2)       = ........ m / s    t = (- 2 yo / a)      = ....... s    the horizontal distance will be    x = vx t         = .........m (b)    so now the supplies have to travel a horizontaldistance of only 425 m    so the time of flight will be    t = 425 m / 69.4 m / s      = ......... s    again using the equaytion of motion we getthe velocity as    vyo = (y -  yo - (1 / 2) ay t2) / t          = ........ m/ s (c)    the vertical speed will be    vy = vyo +ay t         = ......... m /s    so the magnitude of the speed will be    v = (vx2 +vy2)       = ........ m / s

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