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Imagine I kick a ball vertically. What information do I need to determine the velocity of the ball as it leaves my foot, the KE and the GPE at the time the ball is released?

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Imagine I kick a ball vertically. What information do I need to determine the velocity of the ball as it leaves my foot, the KE and the GPE at the time the ball is released?

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sophiecentaur

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Okay, say I have the following problem and I want to calculate the velocity of the ball as soon as it is released and the maximum height it reaches.

If I kick a ball of mass 0.5Kg vertically with a forcé of 30N what will be the ball’s velocity as soon as the ball is released? Assuming the ball leaves my foot at a height of 0.3m from the ground and my foot makes contact with the ball for a distance of 0.05m before releasing it. Gravity = 9.8m/s^2.

Once I have the velocity it should be straight forward to calculate maximum height with V^2=u^2 + 2as, but how do I measure the velocity at the time of release?

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So the acceleration of the ball upwards is 25.1/0.5= 50.2 m/s^2. Use v^2 = u^2 + 2as to calculate the velocity at release. So v^2 = 0+2*50.2*0.05

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Assuming the ball is at rest when you kick it

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sophiecentaur

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So the acceleration of the ball upwards is 25.1/0.5= 50.2 m/s^2. Use v^2 = u^2 + 2as to calculate the velocity at release. So v^2 = 0+2*50.2*0.05

The 'Force' of the kick is not enough to predict how fast the ball will travel. You need to specify

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Also, now that we know all the variables, would GPE and KE (at the time the ball is released) have the following values:

GPE = mgh = 0.5 x 9.81 x 0.3 = 1.47 J

KE = 0.5mv^2 = 0.5 x 0.5 x 5.02 = 1.255 J

Would all that KE be converted to GPE as the ball reaches maximum height?

One more question. Are we exerting a force through a distance every time we are giving motion to an object? Kicking a ball would be a clear example but if we are playing ping pong for instance, are we exerting a force through a distance every time we hit the ball with our table tennis racket? Obviously that distance would be excessively small but it would still not equal 0 would it?

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sophiecentaur

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Air resistance will have an effect on a relative low mass / large area object. You can work out the drag factor for an spherical object in air. Incorporating that would give a better idea of final height. Low velocities will produce less effects from drag, of course.Thanks a lot for the answers! Obviously in practical situations the force would not be constant across that small distance, but if we knew how the force varied and plotted it on a force displacement graph would we still be able to calculate KE (the area under the curve) and therefore the balls velocity when released?

Also, now that we know all the variables, would GPE and KE (at the time the ball is released) have the following values:

GPE = mgh = 0.5 x 9.81 x 0.3 = 1.47 J

KE = 0.5mv^2 = 0.5 x 0.5 x 5.02 = 1.255 J

Would all that KE be converted to GPE as the ball reaches maximum height?

One more question. Are we exerting a force through a distance every time we are giving motion to an object? Kicking a ball would be a clear example but if we are playing ping pong for instance, are we exerting a force through a distance every time we hit the ball with out table tennis racket? Obviously that distance would be excessively small but it would still not equal 0 would it?

Even a table tennis ball impact takes a finite time (so does the impact between the steel balls in a Newton's Cradle). In many calculations about collisions, they use a quantity called Coefficient of Restitution which is the ratio of the separation speed to the approach speed. It's 1 (100%) for a perfect elastic collision and can be anything down to Zero (for two lumps of putty). That is a simplified figure which assumes that the same proportion is lost for all speeds but is pretty successful in practice.

In many collision calculations, the actual time or distance during impact can be ignored and you can just work on the assumption that Momentum is Conserved and deal with the initial and final situations. COR comes in very handy for this, of course. I remember doing endless collision calculations in A Level Applied Maths.

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I see! Thanks a lot for the answer!Air resistance will have an effect on a relative low mass / large area object. You can work out the drag factor for an spherical object in air. Incorporating that would give a better idea of final height. Low velocities will produce less effects from drag, of course.

Even a table tennis ball impact takes a finite time (so does the impact between the steel balls in a Newton's Cradle). In many calculations about collisions, they use a quantity called Coefficient of Restitution which is the ratio of the separation speed to the approach speed. It's 1 (100%) for a perfect elastic collision and can be anything down to Zero (for two lumps of putty). That is a simplified figure which assumes that the same proportion is lost for all speeds but is pretty successful in practice.

In many collision calculations, the actual time or distance during impact can be ignored and you can just work on the assumption that Momentum is Conserved and deal with the initial and final situations. COR comes in very handy for this, of course. I remember doing endless collision calculations in A Level Applied Maths.

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