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If two stones are held level a certain distance above the ground. Stone A is then thrown up and stone B dropped down. Which stone will hit the ground the hardest?
A; A
B; B
C; It depends on how hard you throw stone A
D; It depends on how high above the ground the stones are held.
I'm guessing you're supposed to assume that the stones have the same mass, and the velocities with which they are thrown are equal in magnitude but opposite in direction. In that case, they will hit the ground with the same velocity, or energy, or force, or whatever you want.
Their mass doesn't matter. One is thrown up, you don't know the velocity. The other is dropped.
Do you mean force? force=mass*accQuote:
Which stone will hit the ground the hardest?
In that case, they will have equal force.
However, if you mean inertia(momentum).
momentum=velocity*mass
dist=1/2acc*time^2
Since throwing the stone up in the air would cause it to reach a greater height before freefall, it would take a longer time to hit the ground, and accelerate more, yielding a greater velocity and momentum.
That's assuming their masses are equal.
If you use standard uniform acceleration formulae, it all comes down to whether:
(2h/g)0.5 < u/g + (u2 + 2gh)0.5/g
If it is, B will hit the ground first.
If the two are equal, they will both hit the ground at the same time.
If it's the other way around, A will hit the ground first.
Stick g=10 ish and squaring each:
h/5 < (2u2 + 20h + 2u(u2 + 2gh)0.5)/100
h/5 < h/5 + u2/50 + u(u2 + 2gh)0.5/50
As u must be a positive quantity, it appears that the inequality is true, so B must land before A.
The logic isn't great but to be honest this isn't formal, i don't really give a crap, it just demonstrates the idea.
Well it also appears I answered the wrong question! So:
The dropped stone hits the ground with velocity given by v2 = 2gh
The thrown stone hits the ground with velocity given by v2 = u2 + 2gh
Since u2 must be a positive quantity, the velocity with which the thrown stone hits the ground must be greater than the velocity with which the dropped stone does so.
Better than that, it doesn't matter which direction you throw the stone.
yeah, themanwhocan and digital error, you were both right.
yeah except if you throw it at escape velocity or faster
Thats a very good point, I never though of that.
There's also terminal velocity.Quote:
Originally posted by Acidic
Thats a very good point, I never though of that.
But that, aswell as escape velocity, didn't seem relevant IMO.
I believe the objects need to have equivalent densities also.
{Imagine a pound of feathers vrs a pound of lead. If the feathers aren't compressed down to where they have the same density as lead, then their surface area will slow down their passage.
unless in a vacuum}
Would the answer differ if a small child was used in place of the rock?
:)
lol
Given the same mass/density and outside forces acting on both objects, if the thrown stone was thrown with a magnitude of zero, then they would hit the ground at the same velocity; hence the same force, meaning (edited after I looked at the answers again) that C is correct.
In any case, it is a poorly written question overall.
:)