(adsbygoogle = window.adsbygoogle || []).push({}); 1. The problem statement, all variables and given/known data

A rescue plane wants to drop supplies to isolated mountain climbers on a rocky ridge 235 m below. The plane is traveling horizontally with a speed of 280 km/h (77.8 m/s).

a) How far in advance of the recipients (horizontal distance) must the goods be dropped (Fig. 3 41a)?Already figured this out and got it right: 538.8 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 (Fig. 3 41b)?

Magnitude: ???

2. Relevant equations

y=y0+(vy0)t-(1/2)gt^2

3. The attempt at a solution

Part A work:

y=y0+(vy0)t-(1/2)gt^2

235=(-1/2)(-9.8)(t^2)

t=6.93s

x=x0+vx0t

x=0+(77.8)(6.93)

x= 538.8 m

-------------------------------------

Part B work

d=rt

113.8=77.8t

t=1.5 s

tfinal= 6.93s - 1.5s

tfinal = 5.5s

y=y0+(vy0)t-(1/2)gt^2

235=0+(vy0)(5.5)-(1/2)(9.8)(5.5^2)

vy0~= 70 m/s

I think I did this problem right, but the computer keeps saying that it's wrong. Is this a rounding error? I have tried 70 m/s, but it said that was wrong so I think that I might have to round to 69.7 m/s instead of 70 m/s since the computer is picky sometimes. I only have one chance left to get it right. :P

If this isn't a rounding error, what have I done wrong? Any help is appreciated, thanks.

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