The Student Room Group

Cannon question

Can someone explain why the answer to this question is C?

Im pretty sure its due to the air resistance slowing down the ball when its fired as the velocity fired > falling velocity thus more drag acts on the ball when its going up so it comes to rest quicker - but if thats the case then if the velocity fired < falling velocity then less drag would act, so with that logic shouldnt the answer be D?
image cannon.jpg
I may be completely wrong with my thinking, but any pointers would be greatly appreciated.
(edited 2 months ago)
Reply 1
I could be completely wrong here; physics isn't my strong suit! However, I would've thought that the cannonball would take less time to go up because it has the momentum from, well, being shot out of a cannon lmao. But then when it reaches its peak height, it has 0 velocity, and then starts accelerating until it reaches its terminal velocity. I presume that the time to reach terminal velocity is longer going down than the time to reach 0 velocity going up?
Reply 2
You're probably right i just cant seem to wrap my head around why that happens for some reason, but thanks
Reply 3
Maybe you could try making up values and then putting them into SUVAT when the ball is first shot out, at the peak, and then at its terminal velocity? That could help to see what's going on :wink:
Reply 4
Ordinarily with suvat with no air resistance, the time going up till max height equals time going down, which is why I put my originally wrong answer on the paper.
I dont really know how to implement a increasing factor of drag force alongside suvats - the drag would have to be a function of velocity^2 or smth and im not really too sure how to do that
Reply 5
Original post by mosaurlodon
Ordinarily with suvat with no air resistance, the time going up till max height equals time going down, which is why I put my originally wrong answer on the paper.
I dont really know how to implement a increasing factor of drag force alongside suvats - the drag would have to be a function of velocity^2 or smth and im not really too sure how to do that

The (negative) acceleration on the way up is < -9.8 and on the way down its > -9.8.
So the average speed will be less on the way down and hence the time will be longer to cover the same distance. You dont need an accurate model of air resistance, just that acceleration is < or > -g (and its not necessarily constant either).
(edited 2 months ago)
Reply 6
Ohh I think I get it, so while going up the (downwards) acceleration = weight + drag, while going down the acceleration = weight - drag
and with constant initial velocity going upwards and final velocity while coming down - (as they should be the same?)
If acceleration increases, time must decrease?

So what is the "not necessarily constant" thing you mentioned - are you talking about acceleration?
Reply 7
Original post by mosaurlodon
Ohh I think I get it, so while going up the (downwards) acceleration = weight + drag, while going down the acceleration = weight - drag
and with constant initial velocity going upwards and final velocity while coming down - (as they should be the same?)
If acceleration increases, time must decrease?
So what is the "not necessarily constant" thing you mentioned - are you talking about acceleration?

The "not constant" is just that suvat could give you inspiration (so talk about whether a is less than or greater than g) but its not necessarily accurate depending on how air resistance is modelled.

Tbh, questions like this are less about equations and more about choosing an extreme thought experiment. Imagine firing a guy out of a cannon with their parachute always open (up and down). Theyd rapidly decelerate on the way up and float down gradually. Problem solved.

But youre correct about
net force = gravity +/- drag
and hence reason about the acceleration (but its not really necessary if you pick a good example/experiment).
(edited 2 months ago)
Reply 8
Oh yeah that analogy makes things a lot easier.
Wouldve been good if I thought about that 2 weeks ago.... thanks

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