M1 q.

EDIT: I realised that for some reason i'm assuming that the 15m is its maximum distance, is this why they are different?

A ball is projected upwards from a point which is 4m above the ground with speed 18ms^-1 find:

a)The speed of the ball when it is 15m above its point of projection
b)The speed with which the balls hits the ground

So I drew a diagram and I tried to attempt this question a different way than what the book did:



I ignored the red particle and started from the green:

from the green I said u = 0, a = 9.8 and s = 19 and found the velocity to be 19.29

I then done it another way with u = 18, a = -9.8 and s = -4 and I get the velocity to be about 20ms^-1

these two are very similar so im not sure whats going wrong

EDIT: I realised that for some reason i'm assuming that the 15m is its maximum distance, is this why they are different?

A ball is projected upwards from a point which is 4m above the ground with speed 18ms^-1 find:

a)The speed of the ball when it is 15m above its point of projection
b)The speed with which the balls hits the ground

So I drew a diagram and I tried to attempt this question a different way than what the book did:



I ignored the red particle and started from the green:

from the green I said u = 0, a = 9.8 and s = 19 and found the velocity to be 19.29

I then done it another way with u = 18, a = -9.8 and s = -4 and I get the velocity to be about 20ms^-1

these two are very similar so im not sure whats going wrong

EDIT: I realised that for some reason i'm assuming that the 15m is its maximum distance, is this why they are different?

a) The mistake you made here was to assume that S is the distance from the ground, it isn't, it is the distance travelled.

Actually, both answers were for part (b)... And if something goes up, comes down, and hits the ground, S is the distance from the ground, not distance traveled.

a) The mistake you made here was to assume that S is the distance from the ground, it isn't, it is the displacement.

S= 15
U=18
V=To be found <--Up is positive
A=-9.8
T=Unknown

V^2=U^2+2as
V^2=294
V=17.1ms-1

b)

S=-4
U=18
V=To be found <---upwards is positive
A=-9.8
T=Unknown

V^2=U^2+2as
V^2=402.4
V=20.1ms-1


I hope my calculations are right, I used a pathetic calculator on my computer!

Why did u take a as negative?

same why is a negative if its falling down?

They defined upwards as positive (displacement, velocity, acceleration). Gravity acts downwards so negative -9.8.

then why did they use -9.8 for going upwards when finding the speed?

Going upwards represents the (intial) velocity. It has nothing to do with the way the force (acceleration) acts which downwards.

You'd be better off trying to do the calculation yourself and posting your understanding? A sketch of the trajectory (position or velocity against time) would help and thinking about what features of the graph tell you about the velocity and acceleration.

thanks for your help i think i get it,
for the speed its
s=15,u=18 and a=-9.8 getting 5.5=v
for speed when it hits the ground
its s=-4 ? is it because its below the where it started u=18 and a=-9.8 getting v=20

thanks for your help i think i get it,
for the speed its
s=15,u=18 and a=-9.8 getting 5.5=v
for speed when it hits the ground
its s=-4 ? is it because its below the where it started u=18 and a=-9.8 getting v=20

becuase it says take up positive, if we are going down but not bellow origin would put a negative in the suvat equation or leave it, like for example if it says find speed when it ball drops drops down at 4 metres above the ground

A sketch really does help, marking the positive direction on. All the variables in the suvat have to be consistent with respect to the chosen positive direction. It doesn't matter which direction you choose as positive, but you have to be consistent.

Not sure what your last sentance means. Maybe show your calculation?

the speed it hits the ground means it falls down meaning a positive gravity, why do they use a negative gravity ?

This was explained earlier - the choice of upwards as "positive" means that gravity - which is acting downwards - is automatically "negative" with respect to this choice.

You could in theory choose downwards as the positive direction, but that might make part (a) of the question rather confusing and hard to understand, since initially the ball would be travelling upwards from its start position but its displacement would be negative!

Does your textbook not have some more basic examples of particle motion under gravity that you can use as a basis for tackling questions like these?

ok so they they up is positive so that means the acceleration going up will be -9.8?
and going down will be +9.8?

No!

They choose upwards as positive so acceleration (due to gravity) is always negative.

If you choose downwards as positive, then acceleration is positive.

The sign attached to the force is dependent on your choice of absolute direction, it doesn't flip over just because the ball is moving in a different direction - gravity is always acting in the same direction throughout the motion