Alevel help maths ugentttt

Help I don’t get projectiles…MechanicsYou know if you make the vertical height to be negative (take it as -ve) do you have to take gravity as negative too??
Like for part 3.8…
I took gravity as negative but the 20m as positive and got incorrect ans

(edited 10 months ago)

Reply 1

DFranklin

18

I think "make the vertical height be negative" is a bit simplistic. The key point is to work out what direction you want to measure "y" in (and what your start and end points are).

For the given problem, the following are all fairly reasonable approaches:

Let y be the height above ground, and find out when y = 0. Then y = 20 - gt^2/2 (i.e. gravity provides a negative acceleration, because it's acting downwards, while we're measuring y upwards).

Let y be the distance below from the starting point, and find out when y = 20. Then y = gt^2/2 (now gravity provides a +ve acceleration because it's acting downwards, and we're measuring y downwards).

Let y be the distance above the starting point, and find out when y = -20. Then y = -gt^2/2 (now gravity provides a -ve acceleration because it's acting downwards, and we're measuring y upwards).

(edited 10 months ago)

Reply 2

Alevelhelp.1

OP

15

Original post by DFranklin

I think "make the vertical height be negative" is a bit simplistic. The key point is to work out what direction you want to measure "y" in (and what your start and end points are).

For the given problem, the following are all fairly reasonable approaches:

Let y be the height above ground, and find out when y = 0. Then y = 20 - gt^2/2 (i.e. gravity provides a negative acceleration, because it's acting downwards, while we're measuring y upwards).

Let y be the distance below from the starting point, and find out when y = 20. Then y = gt^2/2 (now gravity provides a +ve acceleration because it's acting downwards, and we're measuring y downwards).

Let y be the distance above the starting point, and find out when y = -20. Then y = -gt^2/2 (now gravity provides a -ve acceleration because it's acting downwards, and we're measuring y upwards).

This concept isn’t sticking in my head and I’ve got my final exam on Monday…
So can I just make gravity negative and distance negative…
and when g is positive distance is positive
Is this the pattern??

Reply 3

mqb2766

21

DFranklin gave a fuller description, but you can always assume that upwards is the positive direction. Therefore because gravity acts downards its always negative acceleration so
a = -g = -9.8

A positive displacement then means upwards from the initial position and a positive velocity means its heading upwards.

A negative displacement means downwards from the initial posiition and a negative velocity means its heading downwards.

Distance and speed are "scalars" so theyre always positive so work out the suvats using the usual "vectors" displacement/velocity/acceleration and then
distance = |displacement|
speed = |velocity|

Choosing upwards as positive isnt always the best approach, but its simple to remember and works and acceleration due to gravity would always be negative.

(edited 10 months ago)

Reply 4

DFranklin

18

Original post by Alevelhelp.1

This concept isn’t sticking in my head and I’ve got my final exam on Monday…
So can I just make gravity negative and distance negative…
and when g is positive distance is positive
Is this the pattern??

No. If it's really too hard for you to read 5 short sentences, you're probably best off doing as mqb suggests: always take upwards to be positive, gravity to be negative. You will still need to think carefully about what your start/end values for y are.

Edit: the downside with that is that if, say, you want to find the time taken for a stone to fall 10m, you need to write y = -gt^2/2 and solve for y = -10, when it's a lot more natural to take y = gt^2/2 and solve for y=10.

(edited 10 months ago)

Alevel help maths ugentttt

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