Question

You are cooking an omelette. An egg (mass 53 grams) slips out of your hand. It hits the ground 0.5 seconds later. From how far up did you drop the egg?

a) From how far up did you drop the egg?
b) How fast if the egg going in the instant before it hits the ground?
c) How long would the egg take to hit the ground if you dropped it from four times higher up?
d) How would your answers for a)-c) change if you dropped a 226 gram ostrich egg?

Thank you!

Answer 1

All answers are given and explained below...

Explanation:

a) If you do not know the speed of the egg the moment it hits the floor, you must use this equation to find the distance it falls:

`d=v_ot+1/2 a t^2`

Since the initial speed is zero when you drop something, the first term of the right vanishes, giving

`d=1/2 a t^2`

where the acceleration `a` is `9.8 m/s^2`

Therefore `d=1/2*(9.8) (0.5)^2= 1.225m`

b) Now for the final speed:

`v=v_o + a*t=(9.8)*0.5 = 4.9 m/s`

c) Go back to the equation of part (a) for this:

`d=v_ot+1/2 a t^2`

Four times higher means `d=4.9 m`, so

`4.9=0+1/2(9.8)t^2`

This solves to give us

`4.9=4.9t^2`, meaning `t^2=1` and `t=1s`

d) None of your answers would be any different if the mass were greater (or smaller) as the acceleration of gravity is independent of mass (especially at these low speeds).

Answer 2

To take the last question first, the mass does not matter when something falls. Explanation below. Answers:
a) `1.225` `m`
b) `4.9` `ms^-1`
c) `1` `s`
d) They wouldn't

Explanation:

If we ignore air resistance, the mass of a falling object does not change its acceleration. The gravitational force on a more-massive object is greater, but its mass is greater by the same amount so the acceleration remains the same.

The initial velocity `u=0` `ms^-1`. To fnd the distance we can use:

`s=ut+1/2at^2=0xx0.5+1/2xx9.8xx(0.5)^2=1.225` `m`

This is the answer to a).

To find the velocity we use:

`v=u+at=0+9.8xx0.5=4.9` `ms^-1`

Four times the height is `4xx1.225=4.9` `m` high.

`s=ut+1/2at^2` but since `u=0`, `s=1/2at^2`

Rearranging to make `t` the subject:

`t=sqrt((2s)/a)=sqrt((2xx4.9)/9.8)=1` `s`

Answer 3

OK, you need a burst of kinematics.

Explanation:

For the first one, use displacement, `s = ut + 1/2a.t^2` because the question says slips (not thrown etc.) we can assume that the initial velocity, u = 0 which simplifies the first equation to `s = 1/2a.t^2` so

I get s = 1.38 m

Next, use `v^2 = u^2 + 2a.s` but again as u = 0 so we can eliminate one term and we get `v = sqrt(2a.s)`

I find v = 5.2 m/s

For part (c) we can just re-use the first equation, with u = 0 and rearrange it for t to give, `t = sqrt((2s)/a)`

My answer is t = 1.06 s but check it!

Finally, (d), the sting in the tail. Panic not, Galileo showed that the acceleration due to gravity is independent of mass, so none of the answers will change at all (assuming air resistance can be ignored, which is reasonable at low velocities.)

The key to all this is (i) knowing or having the kinematic equations in front of you (II) spotting that each one misses out one of the 5 factors (often called s, u, v, a and t in the U.K.) This means if you do a table of the 5 factors you will be missing one you are told to find (the unknown) and one will be irrelevant.

Put in each value to your table with a question mark for the unknown and a dash for the irrelevant one and then it is just algebra and arithmetic to find a solution.

Happy days!