There are about 6 different answers based on how we interpret your question. How much time elapses between releasing an egg and it reaching a distance of 30 feet... How fast is the egg traveling after it's gone 30 feet... What effect does wind resistance have (quantitatively) on the acceleration due to gravity... Ep = MGH Ek = 1/2 MV^2 Ep (start)= Ek (end) therefore MGH = 1/2 MV^2

Weight of One (US) Large Egg: In Shell = 57 grams (doesn't matter because they cancel, but i'll include it here) 30 feet = 9.14400000003658 meters

In the words of MC Front-a-lot, "Have you gathered all the facts that you needed to discern?"

.057KG * 9.81 m/s^2 * 9.144 M == (0.057KG * V^2) / 2 multiply both sides by 2 2 * .057KG * 9.81 m/s^2 * 9.144 M == 0.057KG * V^2 cancel the masses 2 * 9.81m/s^2 *9.144m = V^2 evaluate V^2 = 179.40528 m^2/s^2 square root both sides V = 13.394m/s convert if you really want to be a tool: 13.394 (m / s) = 43.94 feet per second Did I use significant figures there? No, because I simply don't care. :)

Another way to skin the cat: Vf^2 = Vi^2 + 2AD Vf^2 = 0 + 2 *9.81 m/s^2 * 9.144m Vf^2 = 179.40528 Vf = root(179.40528) Vf = 13.394

Whatddaya know, the formulas are really the same! Energy is conserved AND baby jesus doesn't cry about it.

otherwise, d = (Vi + Vf)/2 * T which is handy for solving time, now that we are sure of how fast its going. T = d / (Vi + Vf)/2 T = 9.144m / (0 + 13.394m/s)/2 T = 9.144m / 6.697m/s T = 1.365 seconds.

Now does that logically make sense? acceleration due to gravity is 9.81m/s squared

so in the first of the 1.4 seconds, it travels half the distance, and by the end of that second is going nearly 10m/s. distance travelled is 5ish meters. In that last .4 seconds, its already zooming at 10m/s and speeding up...so its totally reasonable to think it will make the last half of the journey in half the time :)

What will wind resistance be? Negligible, Do as others say and look up blunt body. Long story short it's the square (or cube) of the velocity and the cross section. so with such a small object at such a slow speed in such a slippery fluid (air) -- its really negligible.

I posted the above on the grounds that you read and learn from it. If its homework and you just copy what I wrote down, you'll fail forever and baby jesus will cry.

. Not many ppl around here will do your homework for you, but here's a hint: acceleration due to gravity. And another one: drag coefficient (or coefficient of drag). . Google is your friend.

and use metric! It may be annoying for real-everyday calculations like miles per gallon which are intrinsically more paletable -- but when it comes to physics, use metric. :D

1) 1/2at^2 is the distance, solve for time. the derivative with respect to time is the velocity at a given time into the flight, or a*t 2) negligible for an egg from 30 feet but can be estimated from the blunt body (cross sectional area at its widest point)

## Discussions

7 years ago

There are about 6 different answers based on how we interpret your question. How much time elapses between releasing an egg and it reaching a distance of 30 feet... How fast is the egg traveling after it's gone 30 feet... What effect does wind resistance have (quantitatively) on the acceleration due to gravity...

Ep = MGH

Ek = 1/2 MV^2

Ep (start)= Ek (end)

therefore MGH = 1/2 MV^2

Weight of One (US) Large Egg: In Shell = 57 grams (doesn't matter because they cancel, but i'll include it here)

30 feet = 9.14400000003658 meters

In the words of MC Front-a-lot, "Have you gathered all the facts that you needed to discern?"

.057KG * 9.81 m/s^2 * 9.144 M == (0.057KG * V^2) / 2

multiply both sides by 2

2 * .057KG * 9.81 m/s^2 * 9.144 M == 0.057KG * V^2

cancel the masses

2 * 9.81m/s^2 *9.144m = V^2

evaluate

V^2 = 179.40528 m^2/s^2

square root both sides

V = 13.394m/s

convert if you really want to be a tool:

13.394 (m / s) = 43.94 feet per second

Did I use significant figures there? No, because I simply don't care. :)

Another way to skin the cat:

Vf^2 = Vi^2 + 2AD

Vf^2 = 0 + 2 *9.81 m/s^2 * 9.144m

Vf^2 = 179.40528

Vf = root(179.40528)

Vf = 13.394

Whatddaya know, the formulas are really the same! Energy is conserved AND baby jesus doesn't cry about it.

otherwise, d = (Vi + Vf)/2 * T

which is handy for solving time, now that we are sure of how fast its going.

T = d / (Vi + Vf)/2

T = 9.144m / (0 + 13.394m/s)/2

T = 9.144m / 6.697m/s

T = 1.365 seconds.

Now does that logically make sense?

acceleration due to gravity is 9.81m/s squared

so in the first of the 1.4 seconds, it travels half the distance, and by the end of that second is going nearly 10m/s. distance travelled is 5ish meters. In that last .4 seconds, its already zooming at 10m/s and speeding up...so its totally reasonable to think it will make the last half of the journey in half the time :)

What will wind resistance be? Negligible, Do as others say and look up blunt body. Long story short it's the square (or cube) of the velocity and the cross section. so with such a small object at such a slow speed in such a slippery fluid (air) -- its really negligible.

Answer 7 years ago

http://www.physicsclassroom.com/class/1dkin/u1l6a.cfm

I posted the above on the grounds that you read and learn from it. If its homework and you just copy what I wrote down, you'll fail forever and baby jesus will cry.

Answer 7 years ago

For a second there I thought you were going to explain the archers condumdrum. LOL best answer LOL!!!

Answer 7 years ago

it COULD get to that point :D

7 years ago

. Not many ppl around here will do your homework for you, but here's a hint: acceleration due to gravity. And another one: drag coefficient (or coefficient of drag).

. Google is your friend.

Answer 7 years ago

and use metric! It may be annoying for real-everyday calculations like miles per gallon which are intrinsically more paletable -- but when it comes to physics, use metric. :D

7 years ago

1) 1/2at^2 is the distance, solve for time. the derivative with respect to time is the velocity at a given time into the flight, or a*t

2) negligible for an egg from 30 feet but can be estimated from the blunt body (cross sectional area at its widest point)