# Airplane drops a package (relative motion)

by southernbelle
Tags: airplane, drops, motion, package, relative
 P: 35 1. The problem statement, all variables and given/known data A plane flying at a speed of 275 m/s at an altitude of 3000 m drops a package. The plane doesn't change its direction or speed and there is no wind resistance. a) Where does the package hit the ground relative to the point of release? b) Where does the package hit the ground relative to the moving airplane? 2. Relevant equations R = sqrt A2 + B2 3. The attempt at a solution a. Well I know that it would land directly under it because there is no wind resistance. But I don't know how to come up with that answer. b. I did R = 2752 +30002 and took the square root. I came up with 3,013m.
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P: 5,346
 Quote by southernbelle 1. The problem statement, all variables and given/known data A plane flying at a speed of 275 m/s at an altitude of 3000 m drops a package. The plane doesn't change its direction or speed and there is no wind resistance. a) Where does the package hit the ground relative to the point of release? b) Where does the package hit the ground relative to the moving airplane? 2. Relevant equations R = sqrt A2 + B2 3. The attempt at a solution a. Well I know that it would land directly under it because there is no wind resistance. But I don't know how to come up with that answer. b. I did R = 2752 +30002 and took the square root. I came up with 3,013m.
How long did the package take to hit the ground from 3km?
 P: 35 I did Yf = Yi + Via + 1/2at2 0= 3000 + 0 + -4.9t2 t=25 seconds?
P: 7

## Airplane drops a package (relative motion)

Try;

Time for package to hit the ground.

3000m / -9.8m/s^2 = 306.12s^2 = 17.5s

Position of the plane at t = 17.5s, (time it takes for the package to hit the ground.)

17.5s x 275m/s = 4812.5m

Position of the package at t = 17.5s

Xf = Xi + Vi(t) + 1/2a(t)^2
Xf = 0 + 275/m/s(17.5s) + 0(17.5s)
Xf = 4812.5m

That's how I did it. Hope that helps!
 P: 35 Ok, so 0=3000 + 0 + -4.9t^2 And I got 7.5 seconds?
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P: 5,346
 Quote by southernbelle I did Yf = Yi + Via + 1/2at2 0= 3000 + 0 + -4.9t2 t=25 seconds?
24.74 seconds, mostly close enough.

Now you know how fast it was going horizontally, so given the time and velocity, how far across the ground before the bomb hits? Horizontal speed times time perhaps?
 P: 35 Okay, thanks. But how did you get 3000 divided by 9.8t2? What formula did you plug those numbers in to?
P: 35
 Quote by LowlyPion 24.74 seconds, mostly close enough. Now you know how fast it was going horizontally, so given the time and velocity, how far across the ground before the bomb hits? Horizontal speed times time perhaps?
So it would be 275m/s x 24.74s?
I got 6,803.5

That seems like such a big number and it ends up in m/s2
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P: 5,346
 Quote by southernbelle So it would be 275m/s x 24.74s? I got 6,803.5 That seems like such a big number and it ends up in m/s2
6804 m.

Not the way I look at the units it's not m/s2

Now you have the altitude and the ground distance so you can calculate your displacement as you initially tried.

If the bomb hit on the ground at V*t where is the plane at this instant?

Edit: The number isn't all that far if you consider the speed is over 600 mph.
P: 7
 Quote by southernbelle Okay, thanks. But how did you get 3000 divided by 9.8t2? What formula did you plug those numbers in to?
The package fell 3000m at an acceleration of -9.8m/s^2 with a Vi of 0. Just divide the displacement (3000m) by a = -9,8m/s^2 and it give you the time it took the package to hit the ground in s^2. Then take the square root to obtain s.

d = a x t^2
3000m = -9.8m/s^2 x t^2
3000m / -9.8m/s^2 = t^2
306.12s^2 = t^2 = 17.5s
P: 35
 Quote by LowlyPion 6804 m. Not the way I look at the units it's not m/s2 Now you have the altitude and the ground distance so you can calculate your displacement as you initially tried. If the bomb hit on the ground at V*t where is the plane at this instant? Edit: The number isn't all that far if you consider the speed is over 600 mph.
Okay, thank you!
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P: 5,346
 Quote by eyvhgi557r The package fell 3000m at an acceleration of -9.8m/s^2 with a Vi of 0. Just divide the displacement (3000m) by a = -9,8m/s^2 and it give you the time it took the package to hit the ground in s^2. Then take the square root to obtain s. d = a x t^2 3000m = -9.8m/s^2 x t^2 3000m / -9.8m/s^2 = t^2 306.12s^2 = t^2 = 17.5s
D = 1/2 *a*t2

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