Gravity

[BIGDOGY]

I know objects fall at 32ft/sec/sec, but what is the algerbraic formula to calculate the speed an object will reach when falling from a given height?

[HWKE]

Look at it from the point of view of a missile being projected vertically upwards. Its maximum height is h=1/2gt^2.
So let's say it reached a height of 400 feet.
400=1/2gt^2
800=gt^2
800/32=25=t^2
therefore t=5
So you can see that before we can work out the terminal velocity, we must know the time it will take to travel the distance involved. Ok. Now were coming down.
Terminal velocity =v+gt
=5*32=160ft/sec = about 110mph

Hope this helps

[HWKE]

I was disturbed earlier and couldn't finish the answer.
The algebraic equation is:
Vt=V0+g�√(2h/g) ft/sec
Where
Vt is the terminal velocity
V0 is the initial velocity
h is the height i.e. the distance through which the object falls
g is gravity
If you want to only work in ft/sec, the equation simplifies to:
Vt=32�√(h/16) assuming the object is dropped not propelled downwards.
This takes no account of drag

[mibn2cweus]

Alternatively . . . Final velocity equals the square root of, twice the rate of acceleration times the distance.

Vf = �√ ( 2a * d )

Examples:

Find the final velocity after 100 feet of free fall

a = 32 f/s/s (constant rate of acceleration)
d = 100 ft

√ (64 * 100) = 80 f/s


Find the final velocity after 45 metres of free fall:

a = 10 m/s/s
d = 45 m

√ (20 * 45) = 30 m/s

[shaun_04]

v=u+at

v is the velocity at a given time
u is the initial velocity (if it is dropped this is zero)
a is acc due to gravity 9.8ms^-2
t is the time after the object was dropped.

[THECORBYLOON]

If the measurements are in feet, the speed is eight times the square root of the distance travelled.

[BIGDOGY]

Thanks everybody, thats great!