# Need help on projectile motion by 10:00 tonight (~ 2 hrs 10 mins from now)

1. Nov 13, 2007

### Bensky

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.

2. Nov 13, 2007

### XJellieBX

your values for y and y0 need to be switched. because 0 is not your initial height, 235 is. also, if you designated up as positive and down as negative, your generic equation should be y=y0-(vy0)t-0.5gt^2. try that.

3. Nov 13, 2007

### Bensky

Wow, that was obvious. Thank you, it worked. :)

4. Nov 13, 2007

### XJellieBX

you're welcome =) glad I could help.