Physics

[Draphoelix]

"A person jumps in the air for 0,6 seconds. How high above the ground does he reach?"

Is the answer 0.44 metres?

x = (9,82*0,3^2)/2

 

[Ghan]

Don't you need some kind of initial velocity?

 

[Dave312]

The mid point of the jump will occur at 0.3 seconds. At this point, the persons velocity will be 0 and they will be at the top of their jump. They will also be subject to the force of gravity. This gives you enough information such that you do not need the initial velocity.

You can use the formula h = u*t + 0.5*a*t^2 where u is the initial velocity and h is the maximum height of the jump. Therefore:

h = 0 * 0.3 + 0.5 * 9.82 * 0.3^2 = 0.44m

So yes you are correct :thup:

 

[Hatebreeder]

You could add more factors to it.
Such as the kinetic force (since the force doesn't stay constant) and the relation from weight to acceleration.

But other than that, Dave should have the right solution =)

 

[Jaujarahje]

Heh, just did this stuff in physics. Pretty much you assume air resistance and all that is negligible

 

[BANANAMAN]

I have a question.

How do you calculate the speed and direction of an object after it collides with an object in rest?Assuming the object in rest stays in rest.
And what kind of Collision is this?

 

[Dave312]

I assume you are referring to an elastic collision ie. where both objects do not deform in any way. In these cases, the total momentum (mass x velocity) of the system remains constant. Therefore:

• Initial Momentum = Mass1 * InitialVelocity1 + Mass2 * InitialVelocity2
• Final Momentum = Mass1 * FinallVelocity1 + Mass2 * FinalVelocity2
• Final Momentum = Initial Momentum

Now if the second object's velocity does not change (remains at zero), then the first object will remain at the same speed. The direction at which it travels would depend on the shape of the objects and at which angle they hit each other.