For part b of this question, I used suvat to find the speed of the ball immediately before it hits the ground for the first time, where I made a = g. However, I was supposed to use 0.5g, but I don't know why.

Would anyone be able to help?

Would anyone be able to help?

(edited 1 month ago)

Original post by twisterblade596

For part b of this question, I used suvat to find the speed of the ball immediately before it hits the ground for the first time, where I made a = g. However, I was supposed to use 0.5g, but I don't know why.

Would anyone be able to help?

Would anyone be able to help?

It would help to see what you did for a) (and b) but after the first question part you need to do restitution with the ceiling then descend under gravitational force mg taking air friciton into account (which I suspect is what youre asking about?)

(edited 1 month ago)

Instead, just use the Work Energy Principle Change in Energy = Work Done by Friction/Air Resistance.

This is my answer to part a)

Original post by mqb2766

It would help to see what you did for a) (and b) but after the first question part you need to do restitution with the ceiling then descend under gravitational force mg taking air friciton into account (which I suspect is what youre asking about?)

So because F = ma (Newton's second law), as F = 0.5mg, then a would be 0.5g? Is that why?

Original post by π/2=Σk!/(2k+1)!!

Instead, just use the Work Energy Principle Change in Energy = Work Done by Friction/Air Resistance.

Original post by twisterblade596

So because F = ma (Newton's second law), as F = 0.5mg, then a would be 0.5g? Is that why?

Yes, its newton 2 so

net force = ma = weight - air = mg - mg/2 = mg/2

so a = g/2 (positive down)

(edited 1 month ago)

Original post by mqb2766

Yes, its newton 2 so

net force = ma = weight - air = mg - mg/2 = mg/2

so a = g/2 (positive down)

net force = ma = weight - air = mg - mg/2 = mg/2

so a = g/2 (positive down)

Original post by twisterblade596

I thought I understood but now I don't think I do. So you subtract the weight by the air resistive force to get the new acceleration? But if weight and air resistance are acting in the same direction, then why wouldn't you add the forces together?

In part b) weight is downwards and air resistance is upwards as it acts against the direction of motion (downwards). It makes sense as a parachutist falls slower than a normal person jumping out of a plane as both weights are acting down, but the air resistance/drag recduces its effect.

In part a) both weight and air resitance are down as the direction of motion is up. So a=3g/2 for that question part (positive down again).

(edited 1 month ago)

Original post by mqb2766

In part b) weight is downwards and air resistance is upwards as it acts against the direction of motion (downwards). It makes sense as a parachutist falls slower than a normal person jumping out of a plane as both weights are acting down, but the air resistance/drag recduces its effect.

In part a) both weight and air resitance are down as the direction of motion is up. so a=3g/2 for that question part.

In part a) both weight and air resitance are down as the direction of motion is up. so a=3g/2 for that question part.

Ohh ok I think I get it now. Thank you

Original post by TwisterBlade596

Ohh ok I think I get it now. Thank you

Its a fairly common question type so could easily come up in an exam, so make sure youre clear about which way (air) friction acts which in turn means you have to think about the direction of motion.

Original post by mqb2766

Its a fairly common question type so could easily come up in an exam, so make sure youre clear about which way (air) friction acts which in turn means you have to think about the direction of motion.

Ok will do.

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