A-Level Maths: Q5-01 Projectiles: Introducing Projectiles

[natashaxm]

This has been so useful and I finally feel like I can do projectiles

[thomaswinkworth]

Legend, this is why I always put tlmath infront of my searches ha

[saragarber]

Thank you so much for your thorough explanations. They give me so much clarity. Could you suggest any practice questions?

[rasheede.o]

In a textbook I was using it stating that no external forces act upon the particle that has been projected apart from gravity. But when you project a force aren't you applying a external force to it.

[harryoshea]

Hey Jack,

Thanks for your videos, they are helping me out massively. I just wondered whether the projectile method you use with integration wil recieve marks for Edexcel papers. I used this method because I much prefer it to SUVAT but my teacher said the method was not in the mark scheme and an examiner without a keen eye may award me 0 marks for the method.

How true is this? I cannot seem to find this method on Edexcel mark schemes, but I really like it which is frustrating.

[revan]

How did you integrate velocity to displacement? Specifically the trig bit.

[jazza]

At 4:56 you say 'the initial velocity...' but write it out as v instead of u. Why is that?

[sagamark]

I have gone through all the examples you have given to us there is no example on how to use that formula
So what am asking
Is it inform to use those formulas in exams????

[SadiqunhnabiChoudhury]

Can you please explain this projectile question as the answer makes absolutely no sense to me..
A small ball is projected vertically upwards with speed U m s–1 from a point A that is 12.5 m above horizontal ground. The ball moves freely under gravity until it hits the ground 25/7s later. By modelling the ball as a particle,
(a) find the value of U.
S= ut + 1/2atsquared, but why is it answered using s as -12.5m and t as 25/7s, the time is supposed to be for the whole projectile isn't it, so from a to max point, then to ground?? But max point isnt 12.5m it is a value above that over the ground and why is a negative sign used for s to give the answer as U=14?
I know I'm asking for a big favour, but I am completely confused, it would be very helpful if you reply!

[s.m.]

ai yooo, does anyone know if we need this for OCR MEI B a level

[naomisimmons]

at 6:35, when finding the time it takes to get to the maximum height, why do we only look at usintheta-gt and dismiss the ucostheta?

[ethanteale]

instead of using i and j. Can you use vectors?

[abcxyz-elxk]

Hello, why is there no horizontal acceleration? Surely if something is thrown, it increases speed for some time and then starts to decelerate? Thanks

[mariehill]

If you were just using the vertical component of the object coming downwards, u would be zero, but would v be Usinα or would that be an unknown?