Projectile Motion- Plane releases payload

[Catchingupquickly]

Homework Statement

A pilot is attempting to deliver emergency food and first-aid supplies to an isolated northern community that has suffered severe flooding. The plane has a horizontal velocity of 1.4 10^2 km/h as it flies at an altitude of 1.80 m x 10^2. The community is situated on a dry patch of land that is about 72m x 72m.

a) If the supplies are released just as the plane flies directly overhead, will they touch down on land or in the water? Justify your response with calculations that show exactly where the package will land.

b) When should the supplies be released so that they touch down very close to the centre of the dry patch of land? Answer in terms of distance from the target, not time.

Homework Equations

$\Delta \vec d_v = \frac 1 2 \vec a \Delta t^2$$\Delta \vec d_h = \vec v_h \Delta t$

The Attempt at a Solution

[/B]
Plane is flying at 1.4 x 10^2 km/h = 140 km/h or 38.8 m/s (140km/h x 1000m /3600s)

Altitude is 1.8 x 10^2 meters = 180 meters.

How long until it hits the ground

$\Delta \vec d_v = \frac 1 2 \vec a \Delta t^2$

$180m [down] = \frac 1 2 (-9.8 m/s^2 [down]) (\Delta t^2 \\ \Delta t = {\sqrt {\frac {180m} {4.9 m/s^2}}} \\ \Delta t = 6.1 seconds$

Horizontal distance travelled

$\Delta \vec d_h = (38.8m/s [fwd]) (6.1s) \\ ≈ 237 m$

Where's it going to hit?

72m x 72m = 5184 m

So if released directly overhead that's half-way

5184/2 = 2592 m

So it's going to drift another 237m [forward] therefore it will land at 2592 + 237 = 2829m

b) To release it so it hits dead centre, release at 2592 - 237 or at the 2355th meter.

Critique, feedback, and/or verbal abuse welcomed.

 

[gneill]

Where's it going to hit?

72m x 72m = 5184 m

So if released directly overhead that's half-way

5184/2 = 2592 m

What is this calculation? 72m x 72m would give an area in square meters. The plot of land is a square that is 72 m on each side.

 

[Catchingupquickly]

Ah, yes... I see.

Ok, Plan B

Supplies will fall for 237m and need to hit the half way point of 72m area.

72/2 = 36m

237m - 36m = 201m

The plane should release the supplies 201 meters away from the target area to land in the middle.

Or am I still sniffing glue?

 

[gneill]

That's about right. The "1" is probably unjustified precision; What's the number of significant figures you can claim from your given data?

You should keep a few more decimal places in intermediate values in your calculations, and don't round anything until the end. Otherwise you'll find that rounding and truncation errors will creep into your significant figures when there's more than just one or two calculation steps.

 

[Catchingupquickly]

Thank you for the feedback. The textbook uses the first decimal point in its answers. It's been rounding figures at each step in its examples all the way through so far. So I've been mostly following its example.

 

[gneill]

Thank you for the feedback. The textbook uses the first decimal point in its answers. It's been rounding figures at each step in its examples all the way through so far. So I've been mostly following its example.

Hmph. Not a great example for a textbook to set

 

[Catchingupquickly]

Yeah, it's been rough for a beginning physics student such as myself so far. Most of the lessons have been straightforward, but some of the examples have been explained poorly and some steps have been skipped for brevity. And it only gives one example and asks 1-2 practice questions with answers before it asks a question for submission that is related but may have a twist. And now this significant figures flap... well, thank you for the very useful advice.