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A rescue helicopter wants to drop a package of supplies to isolated mountain climbers on a rocky ridge 200 m below. If the helicopter is traveling horizontally with a speed of 70 m/s (250 km/h), (a) 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 (up or down) so that it arrives precisely at the climbers’ position? (c) With what speed does the package land in the latter case?

Respuesta :

Answer:

a) 447.21m

b) -62.99 m/s

c)94.17 m/s

Explanation:

This situation we can divide in 2 parts:

⇒ Vertical : y =-200 m

y =1/2 at²

-200 = 1/2 *(-9.81)*t²

t= 6.388766 s

⇒Horizontal: Vx = Δx/Δt

Δx = 70 * 6.388766 = 447.21 m

b) ⇒ Horizontal

Vx = Δx/Δt ⇒ 70 = 400 /Δt

Δt= 5.7142857 s

⇒ Vertical:

y = v0t + 1/2 at²

-200 = v(5.7142857) + 1/2 *(-9.81) * 5.7142857²

v0= -7 m/s  ⇒ it's negative because it goes down.

v= v0 +at

v= -7 + (-9.81) * 5.7142857

v= -62.99 m/s

c) √(70² + 62.99²) = 94.17 m/s

The horizontal distance at which the the package must be dropped and the vertical velocity are; 316.23 m and -22.08 m/s

What is the vertical velocity?

A)To fall 200 m, the time required is given by the formula;

t = √(2H/g)

t= √(200/9.8)

t = 4.518 seconds.

The supplies will travel forward by a distance of;

Δx  = 4.518 * 70

Δx  = 316.23 m

B) Time in the horizontal direction is;

t = Δx/V

t = 316.23/70

t = 4.52 s

In the vertical direction;

y = ut + ¹/₂at²

-200 = 4.52u + ¹/₂(-9.81)(4.52)²

100.21 - 200 = 4.52u

u = -99.79/4.52

u = -22.08 m/s

C) Speed at which the package lands in the latter case is;

v = √(70² + (-22.08²))

v = 73.4 m/s

Read more about vertical velocity at; https://brainly.com/question/24949996

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