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 210 km/h (58.3 m/s).

Figure 3-37
(a) How far in advance of the recipients (horizontal distance) must the goods be dropped (Figure 3-37a)?
m
(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 (Figure 3-37b)? (Assume up is positive.)
m/s
(c) With what speed do the supplies land in the latter case?
m/s

Explanation / Answer

For (a), first solve how long it would take a box to drop 235m.

X = v_0 * t + (1/2) * a * t^2

-235 = 0 * t + (1/2)(-9.8m/s^2)t^2
solve for t
t = 6.93 seconds

Now that we have the time it will take to drop, just multiply by horizontal speed to see how far from the drop point you need to be

(a) 6.93 sec * 58.3 m/s = 404.019 meters

For part B. Since it needs to be a farther distance the package needs to fall slower (a longer t) so you infer that there is a slight upward velocity

Solve for falling time t = 425m / 58.3 m = 7.2899 sec

Plug this into a kinematics equation
x = V_o * t + (1/2)* a * t^2
-235 = V_o * 7.2899 + (1/2) * (-9.8) + (7.2899^2)
V_o = 3.484 m/s (note: since this is positive, it is thrown upward. Exactly what we inferred earlier)

C: Vf^2 = Vi^2 + 2ax

Vf^2 = 3.484^2 + 2 * 9.8 * 235

Vf = 67.957 m/s

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