Projectiles 3 • Finding Speed at Given Point • Mech2 Ex6C • 🎲

[@stellafan]

Thank you very much for making all these!!! You really simplify the question and made it way more understandable!

[@BicenMaths]

Thank you! Such a nice thing to say 😃

[@cineacle]

@@BicenMaths why did you need to work out T for the suvat part of the vertical motion q at 11:20 ?

[@jozifff.]

16:30. If you wouldn't mind, how would you do this with upwards positive? When I do it, the u's cancel out so it's clearly wrong

[@BicenMaths]

If upwards were positive, you have: s= -2.5 a= -10 u= 5/13 u t=? and just go from there! :)

[@jozifff.]

@@BicenMaths thats probably where i keep going wrong. how do you know whether to take the positive or negative of acceleration. I always do upwards as positive, so what would acceleration be in terms of + and -? Also, in cases where I'm solving at the maximum height of s and v is 0, what would acceleration be if upwards is positive?

[@BicenMaths]

@@jozifff. If you are taking upwards as positive, then for projectiles question, acceleration will ALWAYS be negative, as it is ALWAYS downwards. If the ball has gone from a cliff top to the floor, then its displacement is also downwards, so s would also be negative. Hope that helps!

[@rid5]

how do you write so neatly with a mouse?

[@BicenMaths]

These are recorded with a very large interactive whiteboard, where the mouse is a physical pen that I could write on the board with! (Which actually means I should be neater!)

[@maygray3253]

Thank you so much sir, really appreciate your amazing explanations. I was just wondering, when going from part a) of the question to b) do we use the rounded answers in a) for the b) or the exact values? Thanks!!

[@BicenMaths]

I would always use the unrounded values as you move through the question, just to ensure you maintain accuracy. If you forget though, mark schemes are usually quite forgiving with things like this as you move through a question :)

[@maygray3253]

Thank you so much!!!

[@ness8007]

11:03 Sir, wouldn't S = -25 because it is projected upwards (as seen from the diagram) and you took downwards as positive ?

[@BicenMaths]

So s is it’s displacement, and this doesn’t have anything to do with the direction it is projected - rather, it measures how far it is at the end of the journey vs the beginning of the journey. In this case, at the end of the journey it is 25m below its starting point, and as this is downwards, we take it is positive. Hope that helps!

[@ness8007]

@@BicenMaths ohh I see thank you so muchh

[@fearbreaker901]

The question at 26:25 , Q7B , how would you know if 52.5 is positive or negative if taking upwards as postive?

[@BicenMaths]

It would have to be going downwards at that point, so would be negative!

[@fearbreaker901]

@@BicenMaths Thank you very much, that makes sense now, keep it up with these videos 😄

[@raghadelrayah6574]

for part c at 19:46 how do you know if we have to look at the horizontal component?

[@BicenMaths]

Because it asks for the speed - not just the vertical speed at that point. This means that we need to combine the horizontal and vertical components, and recombine them to get the overall speed at that point.

[@raghadelrayah6574]

thank you so much @@BicenMaths

[@dydx8407]

for q1 (vector question) part b); what if we took the downward direction as positive, wouldn't the vertical velocity taken become positive and hence it would it be directed upwards as in: u= -24 a=9.81, so v= ( -24) + 9.8x(3) = +5.4 ? or would I have to make v negative like this: (-v)= ( -24) + 9.8x(3) and if so, why would I have to do this since the final velocity v acts in the downwards direction?

[@dydx8407]

actually I just thought about it 🤣, if it was positive when taking the downwards direction as positive, it means its acting downwards. sorry about that

[@BicenMaths]

Don’t apologise! Sometimes writing out our thinking can help us understand better - well done for figuring it out!

[@missachin5054]

Do you have any examples where the particle is thrown at an angle of depression ?

[@BicenMaths]

I'm not sure that I do actually! It'd be just the same, but when you consider the vertical motion, it would - rather than + for the velocity (if you're taking upwards as your positive direction).

[@jingrosie]

I have a question to ask I’m still really confused on how would I know if the distance would be negative or positive due to the positive direction?

[@BicenMaths]

If you set a direction as positive and calculate a negative answer, it is referring to the opposite direction to the one you set as positive, and vice versa for a positive answer!

[@bob820]

In the 2019 paper, it said find the VELOCITY at a given point. I was a bit confused on why they did Pythagoras and put an angle. Is this the best way of doing it?

[@BicenMaths]

Unless the question has been described in terms of vectors, then when it asks for a velocity it wants speed and direction. Not many people got this in 2019 as it was incredibly unclear in the question. If it were in vectors from the start, you could just write it as a vector, or if you defined i and j you could have given it as a vector. Sneaky question that I hope would be better worded in exams this year.

[@mathushamathan1146]

can you use speed=distance/time for every single projectile question when finding the horizontal components

[@BicenMaths]

For horizontal, yes. Not vertical. Vertical is SUVAT.

[@mathushamathan1146]

@@BicenMaths thank you

[@abdisalah8066]

how do you know the speed is 12 at 5:45 but the vertical speed isn't 24?

[@abdisalah8066]

is the horizontal speed always constant

[@BicenMaths]

@@abdisalah8066 Yes, the horizontal speed is always constant, the vertical speed is the one that changes because of gravity! You can tell which are horizontal and vertical from the vector velocity at the beginning.

[@mathushamathan1146]

how do you know u have to do pythag at 24:07

[@BicenMaths]

Because you have horizontal and vertical speed, these are perpendicular, so you can draw as a triangle. Combining them is done via Pythagoras!

[@hafizx9007]

Hello Sir, at 23.00, why cant you just use 'u' for speed, the 'u' you found in part (a) question.

[@BicenMaths]

This is because the speed is a combination for the horizontal and vertical components - so you need to combine both of them using Pythagoras. The vertical speed is changing, which is why we’re calculating it here using SUVAT.

[@hafizx9007]

@@BicenMaths Oh, ok thank you sir.

[@hamzahaldabbagh4439]

Hi sir why could we not use time in part C even tho its universal?

[@BicenMaths]

I’m not quite sure what you mean, but you could find the time it is 12.4m above the ground and use this to find the speed, but it’s not necessary to know the time to know the vertical speed, just the initial speed, the acceleration and the displacement!

[@felixjoy1780]

is the horizontal speed always const ant?

[@BicenMaths]

Yes! Unless there is air resistance (which there won’t be for A-Level Maths).