5. A rescue helicopter wants to drop a package of supplies to isolated mountain

Question

5. A rescue helicopter wants to drop a package of supplies to isolated mountain climbers on a rocky ridge 200. m below.

a. If the helicopter is traveling horizontally with a speed of 70.0 m/s, how far in advance of the recipients (horizontal distance) must the package be dropped?

b. Suppose, instead, that the helicopter releases the package a horizontal distance of 400. m in advance of the mountain climbers. What vertical velocity should the package be given so that it arrives precisely at the climbers’ position?

C. With what speed does the package hit the ground in Part b?

Explanation / Answer

A) time taken by packet to drop = sqrt (2h/g)

= sqrt (2*200/9.8) = 6.389 s

Distance =vt =70*6.389= 447.23 m

B) time taken = 400/70 = 5.714 s

Using second equation of motion,

H = ut + 0.5gt^2

200 = 5.714u + 4.9*5.714^2

u = (200-4.9*5.714^2)/5.714

= 7 m/s answer

C) vertical striking speed = 7+5.714*9.8 = 63 m/s

Speed = sqrt (63^2 +70^2)

= 94.2 m/s answer

Related Questions

Navigate

• Browse (All)
• Browse 5
• Subjects

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