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SUVAT Question

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?Screenshot 2024-05-16 214354.png
(edited 1 month ago)
Reply 1
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?Screenshot 2024-05-16 214354.png

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.
IMG_5745.jpg 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.
That was another option but I preferred the SUVAT method.
Reply 6
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)
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?
Reply 8
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.

Ohh ok I think I get it now. Thank you :smile:
Original post by TwisterBlade596
Ohh ok I think I get it now. Thank you :smile:

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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