Projectiles : Above horizontal at a height : Tutorial 2 : ExamSolutions

[@pinklemonade6597]

10 years later and these videos are just as useful they’ve helped me so much thank you!!

[@ExamSolutions_Maths]

They are standard formulae that you should already be familiar with on constant acceleration. My website has tutorials on these. Just check out the index. Sorry cannot post a link here as KZbin prevents it.

[@kyliek.9481]

ExamSolutions hi! Can you please make more Edexcel m2 past paper tutorials if possible? Thank you for these videos btw!

[@ethanroee]

Going through these videos before my a2 maths mechanics exam tomorrow makes a huge difference, you are a legend! Edit: well it’s four years later or so, what a throwback lol. I did end up passing that exam, went to uni got a first in computer science and just got my first job. How things work out ey. Oh and there’s a bloody pandemic

[@ExamSolutions_Maths]

I hope all goes well for you.

[@pinklemonade6597]

@@Name7.62 good luck to you!!

[@willlamble5630]

Thanks alot! i didnt rellay understand projectiles before i saw your videos! cheers.

[@mich3lleeee]

where did you get all these formulas from? is there a list?

[@arizh4033]

Thank you so much sir My classmates and I are so grateful

[@mryonzuna7837]

Thanks for this one

[@DahliaArgue]

thank you

[@ExamSolutions_Maths]

No probs, Hope my site proves useful.

[@elliotfryatt454]

I’m confused with your rearranging on the max height section. You say to move 2(9.8) to both sides and then divide by 2(9.8). So you want me to divide 2(9.8) by itself?

[@ExamSolutions_Maths]

We skipped a small step there so I see your confusion. Since a = -9.8, we have 0 = (8.452..)^2 - 2(9.8)h. Then basically what's meant on the video is to make h the subject, so here's the rearrangement. h = (8.452..)^2 / 2(9.8) . That should give you the answer for h. Does that help?

[@EpicEditsBySaz]

I won't be able to see my teacher before Monday, and this video is bang on the nail, thanks! I was a bit confused on what is negative and positive.

[@ExamSolutions_Maths]

I'm pleased to hear you have it sorted. Good luck if you are taking exams on Monday.

[@karimmagdy6625]

i don't understand how the time of flight is calculated as if the particle is projected with horizontal speed only ?? i mean why did you neglect the time it will take to reach the maximum height and add it to the time you calculated ??? thanks

[@ExamSolutions_Maths]

It takes the same time to go at the constant horizontal component of speed as it takes to go up and down with the vertical component of speed. Working in this way breaks down the calculation into a simpler problem.

[@karimmagdy6625]

I am sorry, i don't get it.

[@ghoda-r3854]

will the final velocity at the bottom (at -10 displacement) be negative or positive?. using your values ie s=-10,u=8.452, a =-9.81,t= 2.5 i get both positive and negative values for final velocity from different equations. v=u+at gives me a neg ans while v^2=u^2+2as give me a positive??

[@SamuelOwusu01]

Thank you so much for this video

[@sharmisthatripathy9623]

In the 1st video v was -ve but in this video it’s unknown . I am not understanding that part .

[@stephanieokobi6122]

in the time of flight part, I used the formula for the time of flight and I didn't get the answer you got, why?

[@varunabishek5364]

same here

[@Dawoodj736]

I don't understand why v is unknown, why is it not -20sin25? Please could you get back to me on this. Thank you for your videos btw, they help a lot!

[@ExamSolutions_Maths]

+dawood_jay The vertical velocity is always changing. At the start it is -20sin 25 but at the bottom when the particle has travelled up then down to the ground it will have changed.

[@ubaidsyed5475]

ExamSolutions in tutorial 1 you said that V is the negative of the vertical component but but in this tutorial it's not and I don't understand the difference???could you please explain

[@bushranaheed7531]

I think it is because the ball does not land from the same distance of where it was projected. In the first tutorial, it was projected from the one point and it landed on the same point but further away.

[@leeparker9095]

I don't understand how you got the value for t to be 2.5s using h=10m when the total height is 13.6m.

[@ExamSolutions_Maths]

It has nothing to do with the max height. s is displacement and it was set to -10 not 10 so it gives the time when the particle hits the ground as it is 10m below the initial position.

[@coldbrewedicecoffee6319]

how comes the final velocity is unknown and not -20sin25?? Please help

[@Dan-gh1sd]

NinthNinjaHD because that would be if the horizontal direction is in a straight line. V in this case also has to drop 10 m as well as -u, hope that helps? If not just ask and I’ll try give a clearer explanation

[@Xfrome]

Surely for the 'time of flight' part you've only taken into account the time after the particle has gone up, down, and reached 10m above ground once more? What about the initial motion??

[@ExamSolutions_Maths]

No I haven't. I took s (the displacement) to be -10 which represents 10m below the point of projection. If I wanted the time that you suggested then s would equal 0. Hope you follow?

[@Xfrome]

Of course, I didn't think that through at all. Much appreciated!

[@ExamSolutions_Maths]

Tom B-Wake You're welcome