Thread: accelaration

Hi Guys,

If I know the mass of something and know that i want to to attain a certain speed (lets say 3 metres per second) when it hits the ground (assuming acceleration due to gravity as being 9.81 ms^-2

How would I go about finding out at what height the object should be dropped?

(ignoring air resistence)

Don't know if I am missing something?



s = the distance between initial and final positions (displacement) (sometimes denoted R or x)
u = the initial velocity (speed in a given direction)
v = the final velocity
a = the constant acceleration
t = the time taken to move from the initial state to the final state

Replace with . The equations will work with any mass falling to earth.

Hi, thanks,

Trouble is, I dont know from what height the object is to be dropped, nor do i know the length of time it is allowed to drop. :s

All I know is the acceleration due to gravity is 9.81 ms^2

And the said object is 3.1kg

and it needs to attain a speed (once dropped from "x" height) of 5.42 ms^-1.

Originally Posted by KALSTER

Don't know if I am missing something?



s = the distance between initial and final positions (displacement) (sometimes denoted R or x)
u = the initial velocity (speed in a given direction)
v = the final velocity
a = the constant acceleration
t = the time taken to move from the initial state to the final state

Replace with . The equations will work with any mass falling to earth.

Use the forth one. You have all the variables except for displacement.

Originally Posted by KALSTER

Use the forth one. You have all the variables except for displacement.

aha, makes sense thanks,

Another quickie....... I know that F = m x g

But how would i calculate the force of something travelling at say 20 ms^-1 if the mass is say 10 kg.

Because if i were to just use the old F = m x g then I would calculate.....

F = 10 x 9.81

Force = 98.10 Newtons. But im sure im supposed to use 20 ms^-1 somewhere but not too sure where?

Is this homework?

Anyway, in which direction is that ?

If it is in the downwards direction, then just add it to the final velocity. If it is perpendicular to the force of gravity, then it would not affect the equation. The object would still hit the earth at the same time as it would have if it was just falling straight.

What is it specifically that you want to work out?

Originally Posted by KALSTER

Is this homework?

Anyway, in which direction is that ?

If it is in the downwards direction, then just add it to the final velocity. If it is perpendicular to the force of gravity, then it would not affect the equation. The object would still hit the earth at the same time as it would have if it was just falling straight.

What is it specifically that you want to work out?

I am trying to work out the force that 3.1kg will experience when it hits the ground at a speed of 5.42 ms^-1

but F = m x g doesnt seem to take into account that the object would be travelling at 5.42 ms^-1

Common sense would dictate that i could use the formula F = m x v. but something tells me this isnt quite right ?

Its okay, i knew there was a variable missing somewhere !! :-)

Got it

What did you figure out? I was thinking of impulse. :?

Originally Posted by KALSTER

What did you figure out? I was thinking of impulse. :?

I failed to mention (and take into account) the rate of deceleration which is 250g

therfore i CAN use F = m x g

(but i dont use 9.81 as "g")

I use 250 * 9.81 as "g"

Thats where i kept going wrong with it, but i got it now XD

Thanks

Ah, ok. would be then in this context.

So, was it homework or something?

As KALSTER mentioned, force doesn't make sense in this frame (the force in a collision is HUGE, but only for a split second). What you want is impulse.

The equation is basically mass * change in velocity. If the collision is inelastic (your mass "sticks" into the ground without bouncing) then impulse is simply mass * velocity at impact.

If you want to know how much energy is produced from the collision, you would just take the kinetic energy of the mass before collision (1/2 * mass * velocity ^ 2). If the collision is elastic and bouncy, it's a little more complicated. Check out the coefficient of restitution.