Bensky
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Homework Statement
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: ?
Homework Equations
y=y0+(vy0)t-(1/2)gt^2
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
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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.