Also got lost on this question: A rescue plane wants to drop supplies to isolated mountain climbers on a rocky ridge 235m below. If the plane is traveling horizontally with a speed of 250 km/h (69.4 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 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. (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 235m below. If the plane is traveling horizontally with a speed of 250km/h (69.4 m/s), (a) how far in advance of the recipients (horizontal...

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aristocles aristocles · Answer 2 · 2019-10-16T03:49:56+02:00

Answer:

Part a)

[tex]d = 480.4 m[/tex]

Part b)

[tex]v_y = 8.41 s[/tex]

Explanation:

Part a)

Since we know that the vertical speed of the plane is zero

so here the mass is dropped down with zero initial speed

so we will have

[tex]y = \frac{1}{2}gt^2[/tex]

[tex]235 = \frac{1}{2}(9.81)t^2[/tex]

[tex]t = 6.92 s[/tex]

So the horizontal distance moved by it is given as

[tex]d = vt[/tex]

[tex]d = (69.4)(6.92)[/tex]

[tex]d = 480.4 m[/tex]

Part b)

If plane drops the good at distance of 425 m

so the time taken by it to reach is given as

[tex]t = \frac{d}{v}[/tex]

[tex]t = \frac{425}{69.4}[/tex]

[tex]t = 6.12 s[/tex]

now in vertical direction

[tex]y = v_y t + \frac{1}{2}g t^2[/tex]

[tex]235 = v_y (6.12) + \frac{1}{2}(9.8)(6.12^2)[/tex]

[tex]v_y = 8.41 s[/tex]