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.

Part A

If the plane is traveling horizontally with a speed of 217 km/h (60.3 m/s ), how far in advance of the recipients (horizontal distance) must the goods be dropped (see the figure (Figure 1) (a))?

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 (see the figure (b))?

Part C

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

A rescue plane wants to drop supplies to isolated mountain climbers on a rocky ridge 235 m below.

Part A

If the plane is traveling horizontally with a speed of 217 km/h (60.3 m/s ), how far in advance of the recipients (horizontal distance) must the goods be dropped (see the figure (Figure 1) (a))?

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 (see the figure (b))?

Part C

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

Explanation / Answer

A)
y = 235m
Viy = 0
Vix =60.3m/s
x = ?
ay = -9.81m/s^2
ax = 0 (assume negligible wind resistance)

Find the time it takes for the package to fall the 235m to the ground, where it reaches the climbers
y = (Viy)t + (1/2)ay(t^2)
235m = (0m/s)(t) + (1/2)(-9.81m/s^2)t^2
235m = (-4.905m/s^2)t^2
47.9s^2 = t^2
6.92s = t

Now, use that value along with the other givens to solve for x
x = (Vix)t + (1/2)ax(t^2)
x = (60.3m/s)(6.92s) + (1/2)(0m/s^2)(6.92s)^2
x = 417 m

B)
Givens:
x = 425m
y = 235m
Vix = 60.3m/s
Viy = ?
ay = -9.81m/s^2
ax = 0
t = ?

This time, it's the reverse. Use the equation in the x-dimension to solve for t, then use that in the y-equation to solve for Viy
x = (Vix)t + (1/2)ax(t^2)
425m = (60.3m/s)t + (1/2)(0m/s^s)t^2
7.04s = t

y = (Viy)t + (1/2)ay(t^2)
235m = (Viy)(7.04s) + (1/2)(-9.81m/s^2)(7.04s)^2
235m = (Viy)(7.04s) - 243m
478 m = (Viy)(7.04s)
67.91m/s = Viy

C) The horizontal component of the speed is of course 60.3m/s, same as when they're dropped.
Vfy = Viy + (ay)t
Vfy = 67.91m/s + (-9.81m/s^2)(7.04s)
Vfy = -1.15 m/s

The total speed of the supplies can be found using the Pythagorean theorem
Vf = SQRT(Vfx^2 + Vfy^2)
Vf = SQRT[(60.3m/s)^2 + (-1.15m/s)^2]
Vf = 60.31m/s

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