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

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Homework Help Overview

The problem involves projectile motion, specifically the scenario of a rescue plane dropping supplies to mountain climbers located 235 m below. The plane travels horizontally at a speed of 280 km/h, and the discussion focuses on calculating the necessary horizontal distance for the drop and the required vertical velocity for the supplies to reach the climbers.

Discussion Character

  • Exploratory, Assumption checking, Problem interpretation

Approaches and Questions Raised

  • The original poster attempts to solve for the vertical velocity needed for the supplies to reach the climbers after dropping them from a horizontal distance of 425 m. They express uncertainty about whether their answer is correct due to potential rounding issues.

Discussion Status

Some participants provide feedback on the original poster's calculations, suggesting that there may be an error in the assignment of initial and final heights in the equations used. This indicates an ongoing exploration of the problem setup and the equations involved.

Contextual Notes

The original poster mentions a time constraint for submitting their work, which adds urgency to the discussion. There is also a note about the computer's sensitivity to rounding, which may affect the final answer.

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).
3_41.gif


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
-------------------------------------
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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y=y0+(vy0)t-(1/2)gt^2
235=0+(vy0)(5.5)-(1/2)(9.8)(5.5^2)
vy0~= 70 m/s

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.
 
XJellieBX said:
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.

Wow, that was obvious. Thank you, it worked. :)
 
you're welcome =) glad I could help.
 

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