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 H = 235 m below. If the plane is traveling horizontally with a speed of 207 km/hr (57.5 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 x = 421 m in advance of the mountain climbers. What vertical velocity (use the positive direction as upwards) should the supplies be given so that they arrive precisely at the climbers' position? c)With what speed do the supplies land in the latter case?

Explanation / Answer

put ur values in example

A)
?y = 235m
Viy = 0
Vix = 79.2m/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 = (79.2m/s)(6.92s) + (1/2)(0m/s^2)(6.92s)^2
?x = 548m

B)
Givens:
?x = 425m
?y = 235m
Vix = 79.2m/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 = (79.2m/s)t + (1/2)(0m/s^s)t^2
5.37s = t

?y = (Viy)t + (1/2)ay(t^2)
235m = (Viy)(5.37s) + (1/2)(-9.81m/s^2)(5.37s)^2
235m = (Viy)(5.37s) - 141.4m
376.4m = (Viy)(5.37s)
70.1m/s = Viy

C) The horizontal component of the speed is of course 79.2m/s, same as when they're dropped.
Vfy = Viy + (ay)t
Vfy = 70.1m/s + (-9.81m/s^2)(5.37s)
Vfy = 17.42m/s

The total speed of the supplies can be found using the Pythagorean theorem
Vf = SQRT(Vfx^2 + Vfy^2)
Vf = SQRT[(79.2m/s)^2 + (17.42m/s)^2]
Vf = 81.1m/s

Related Questions

Navigate

• Browse (All)
• Browse A
• Subjects

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.