I must be stupid

[Zack]

I can't figure this out. I've thought and thought and looked all around the internet, but I can't find anything that would help me solve a problem I though of...
Discarding air resistance and the fact that the force of gravity decreases with distance (we are dealing with small scales here), and given that an object accelerates towards Earth at 9.8 m/s/s, how do we figure out what its velocity is if it started falling at any given altitude?
So if a 5-gram raindrop falls from an altitude of 1 km, how fast will it be going when it hits Burt Bacharach's head? We can figure it easily if we know how long it took to fall, but we don't, so...how?
This is probably obvious, but I'm struggling with it for some reason.

[Ryan]

Is he wearing a hat?

[RyanKelly]

It has been some time since I've worked with calculus. If anyone spots an error here, please point it out.

You know the following: the rate of acceleration, the initial velocity, and the distance travelled before impact.

Your unknown is the amount of time it will take. The velocity at impact is a function of the time, so this is merely one unknown.

Using the rate of acceleration, set up an equation for the velocity.

Your mass is only needed to calculate the force with which it will hit Burt's head (or if you want a better approximation of the rate of acceleration).

velocity=9.8*time

Integrate over t=0 to x to find

distance=4.9x^2

x is our unknown time.

Solve for x

x=sqr(1000/4.9)

Solve for unknown velocity

velocity= 9.8*sqr(1000/4.9)

140 meters per second.

Well, actually velocity=9.8*sqr((1000-HeightofBurtinMeters)/4.9)

[na_th_an]

v = t * a

isn't it?

In the real world, the raindrop will get decelerated by friction with the athmosphere, so it will reach a maximum velocity and won't accelerate more.

[Zack]

RyanKelly: how did you get to the second equation, distance=4.9*x^2?
Nathan: yes, that's called the terminal velocity. It will reach a certain speed and stop acceleration. But I want to discard air resistance.

[RyanKelly]

v = t * a
isn't it?

To solve for v, we need to know t, which we don't.
We need to find t in terms of the distance.

RyanKelly: how did you get to the second equation, distance=4.9*x^2?

I needed an equation that relates the distance to the time.
The the equation representing velocity in terms of time is the first derivative of the equation representing displacement or distance in terms of time.
The integral of the first derivative of a function is identical to that function, so I merely reversed the standard method of derivation.

y=a*x^b
dy/dx=b*a*x^(b-1)

y=4.9*x^2
dy/dx=9.8*x^1 = 9.8*x

Here b*a=9.8 and b-1=1, thus b=2 and a=9.8/2=4.9

[Zack]

Oh dear. I haven't done calculus yet.
I'll read up on it and get back to you.

[RyanKelly]

Your problem would have stumped me without calculus.
The key is merely to appreciate the significance of the "slope" or "instantanious velocity" of a function. Beyond that, its just a handful of methods for quickly calculating them.... with the execption of those damned trig identities and their mind numbing complex equivalents.

[Antoni Gual]

y=1/2*a* t^2
t=sqrt(2* y/a)=sqrt(2*1000/9.81)=14.27 sec
v=t *a =14.27*9.81=140 m/s *3.6 =504 Km/h

Mr Bacharach should wear a helmet...

[Rattrapmax6]

y=1/2*a* t^2
t=sqrt(2* y/a)=sqrt(2*1000/9.81)=14.27 sec
v=t *a =14.27*9.81=140 m/s *3.6 =504 Km/h

Mr Bacharach should wear a helmet...

Helmet? Not really in real life, somewhere in Physics I read anything faster than 200Km/h avearge is impossible for a free falling object on Earth.... Otherwise, as it cared to point out, Hail balls would turn into Cannon balls and we poor humanbeings/animals on earth would be screwed in any normal shelter like a house, etc.... Something to do w/ compression, air resistance, etc. that acts as a speed-governor...

Been awhile since I read it tho.... :roll: