Idk why I can't do this easy projectile motion question (1 Viewer)

[OreoMcFlurry]

Hey guys,

I was just doing the 2021 NEAP physics paper and I'm kinda confused about this question:


I know that if there are 5 frames per second - this means that the interval between each frame is 0.2 seconds and since the first frame is just the starting frame, then the time of flight is:
14 x 0.2 = 2.8 seconds right?

But it's solving for the maximum height which I struggle with most. I know that for max height, v(y) = 0

I'm not too sure as to what to actually do and so could someone please help out with solving this. It would be highly appreciated - thank you.

 

[notme123]

I struggled with this as well but the main thing is that the time it takes to fall helps determine the maximum height. Since all objects fall at the same rate, just knowing the time it takes to fall is enough to give vertical distance traveled. So here, it takes 1.8s to fall.
y=1/2gt^2
t=1.8
y=15.876 m

 

[OreoMcFlurry]

I struggled with this as well but the main thing is that the time it takes to fall helps determine the maximum height. Since all objects fall at the same rate, just knowing the time it takes to fall is enough to give horizontal distance traveled. So here, it takes 1.8s to fall.
y=1/2gt^2
t=1.8
y=15.876 m

Thank you so much! This makes a lot of sense now that I think about it as max height is measured from the ground to the max height it reaches. I just kept trying to find the initial vertical velocity and tried to use the way up to find max height - but that's not really the max height as it's only measuring from the person to the max height it reaches. (green arrow is the correct one lol)




Nonetheless, I understand where I went wrong lol - thank you very much for the help

 

[yashbb]

Wait what how did u get 1.8?

 

[ExtremelyBoredUser]

Wait what how did u get 1.8?

Every frame, position of the ball, is 0.2 seconds since the camera's 5 frames per second as stated. If you count the total frames, theres 14 so total time is 2.8 seconds however to find max height, we're only trying to find the time taken on the left side of the parabolic trajectory so we deduct the frames right of the highest point, hence 2.8 - 5*0.2 = 2.8 - 1 = 1.8s

 

[yashbb]

Every frame, position of the ball, is 0.2 seconds since the camera's 5 frames per second as stated. If you count the total frames, theres 14 so total time is 2.8 seconds however to find max height, we're only trying to find the time taken on the left side of the parabolic trajectory so we deduct the frames right of the highest point, hence 2.8 - 5*0.2 = 2.8 - 1 = 1.8s

Im going to ask a very dumb q, why is it 14 fps

 

[yashbb]

Hey guys,

I was just doing the 2021 NEAP physics paper and I'm kinda confused about this question:
View attachment 32692

I know that if there are 5 frames per second - this means that the interval between each frame is 0.2 seconds and since the first frame is just the starting frame, then the time of flight is:
14 x 0.2 = 2.8 seconds right?

But it's solving for the maximum height which I struggle with most. I know that for max height, v(y) = 0

I'm not too sure as to what to actually do and so could someone please help out with solving this. It would be highly appreciated - thank you.

Would you mind sending the paper

 

[yashbb]

tysm all good for the answers

 

[windebygirl]

Sure, unfortunately I don't have the solutions as my teacher hasn't sent me that, but here's the paper:

(once I get a hold of the answers, I'll post them up here)

i don't have the whole paper's solutions but this question was in my trial and here's the answer

 

[yashbb]

i don't have the whole paper's solutions but this question was in my trial and here's the answer

Thank you!

 

[CM_Tutor]

I've had to delete the post in this thread in which an upload of a copyright-protected NEAP paper was provided. I have also recently had to delete a copy of an HSC paper from the Notes and Resources as NESA-owned materials cannot be hosted by others without their express permission.

Please, do not upload such materials. If an organisation whose paper was uploaded could show that BoS hosted such materials and that BoS admins / moderators took no action about it, the company that hosts this website could be sued. The Moderator team and BoS admin thus have no choice but to remove such materials as soon as they are seen... if we didn't, the worst-case scenario would include BoS being shut down, which none of us wants. It is for the same reason that THSC cannot host such materials Thank you for cooperating when we take steps that are necessary to protect BoS and ensure that this great site remains available for current and future students.

 

[CM_Tutor]

Question... does the last part of the solution seem strange to anyone else? From the diagram, it appears to me that the maximum height occurs at one second after the ball is thrown.

 

[OreoMcFlurry]

I've had to delete the post in this thread in which an upload of a copyright-protected NEAP paper was provided. I have also recently had to delete a copy of an HSC paper from the Notes and Resources as NESA-owned materials cannot be hosted by others without their express permission.

Please, do not upload such materials. If an organisation whose paper was uploaded could show that BoS hosted such materials and that BoS admins / moderators took no action about it, the company that hosts this website could be sued. The Moderator team and BoS admin thus have no choice but to remove such materials as soon as they are seen... if we didn't, the worst-case scenario would include BoS being shut down, which none of us wants. It is for the same reason that THSC cannot host such materials Thank you for cooperating when we take steps that are necessary to protect BoS and ensure that this great site remains available for current and future students.

Very sorry about this, I completely understand and I will keep this in mind for next time

 

[username_2]

Question... does the last part of the solution seem strange to anyone else? From the diagram, it appears to me that the maximum height occurs at one second after the ball is thrown.

It doesn't make sense because the trajectory is not symmetric. Furthermore, they are sort of assuming that there is no horizontal velocity and hence, would result in an improper answer. But most of all, really, is the value of t they are using for the second part when compared to the first part. (2nd part is half of the time of flight whereas the value used in part one is 9 frames -> 1.8 which is clearly not 2.8/2. Bad question but good idea - they were trying to confuse the students... i think. I would never have gotten this in a test, would've just skipped it.