M1 acceleration (help!)

v = u + at and s = ut + 0.5at^2

Reply 2

What's mass got to do with it?

Reply 3

rpan161

OP

Not sure lol, its just there.

How would I start the question out? what is the time? t=0?

Reply 4

It's got to fall by 20j.

If g=10 this takes 2 seconds.

Position must be 10i+30k+2(20i+50k) = 50i+130k

Which board is this M1?

Reply 5

ghostwalker

Firstly, you need to work out the time of the flight.

This is going to be determined by the vertical components of the position, and velocity; traditionally this is the components in the k direction.

So use suvat (or whatever acronym you've been taught) on the vertical component.

Reply 6

ghostwalker

Firstly, you need to work out the time of the flight.

This is going to be determined by the vertical components of the position, and velocity; traditionally this is the components in the k direction.

So use suvat (or whatever acronym you've been taught) on the vertical component.

I thought the vertical component was j.

Reply 7

ghostwalker

steve2005

I thought the vertical component was j.

If you just have two co-ordinates and one is for the vertical, then it's j.

With 3 co-ordinates it's the k, in my experience, but, this is only tradition, and a given question could specify any as being the vertical.

Edit: Brain dead as usual! There are good reasons for the k direction to be the vertical, which will come out when/if you do vector products.

Reply 8

Doesn't this make the question, or part question, under consideration potentially ambiguous?

Reply 9

ghostwalker

If you just have two co-ordinates and one is for the vertical, then it's j.

With 3 co-ordinates it's the k, in my experience, but, this is only tradition, and a given question could specify any as being the vertical.

OK thanks.

Reply 10

ghostwalker

sciortino

Doesn't this make the question, or part question, under consideration potentially ambiguous?

Not as long as everyong is aware of the convention; which they clearly aren't.

What's iffy to me is that there are no units specified in the question, and if we take the unit vectors to be metres, then someone has a VERY powerful throwing arm.

Reply 11

ysna1

rpan161

A ball of mass m=20 is thrown with initial velocity v=20i+50k from 10i+20j+30k at time t=0. Only gravitational force acts on the ball. Where does the ball land on ground?

Which formula/s would i use to solve this?

There is no vertical speed if you like in the initial velocity.
Hence ball is thrown horizontally.
so for vertical motion
u=0
a=9.8m/s^2
s=20

using s=ut + 0.5at^2
=>20=4.9t^2 thus it takes 10(2)^0.5/7s to reach the floor ten root 2 over 7..

considering the k and i vectors
the speed of projection in the k vector is 50m/s
using distance = speed x time
=>d=50x(10(2)^0.5/7) thus the ball travels 101m in the k vector

similary
for i vectors
d=20x(10(2)^0.5/7) thus the ball travels 40.41in the i vector

for the j vector, the ball lands on the ground so it has no height

hence we get final answer of
101i + 0j +40.41k +10i + 30k= 111i + 0i + 70.41K
i think..

Reply 12

Sounds like a very un-M1 question to me!

Reply 13

ysna1

to me it seems more like a M2 question..

Reply 14

Quite so.
I still don't see the need for mass.
Also, we have no units.
The question would have to confirm the directional convention in use.

Reply 15

ysna1

mass does not matter unless there is an another part to this question..
for me i is always horizontal and j is always vertical

Reply 16

ysna1

There is no vertical speed if you like in the initial velocity.
Hence ball is thrown horizontally.
so for vertical motion
u=0
a=9.8m/s^2
s=20

using s=ut + 0.5at^2
=>20=4.9t^2 thus it takes 10(2)^0.5/7s to reach the floor ten root 2 over 7..

considering the k and i vectors
the speed of projection in the k vector is 50m/s
using distance = speed x time
=>d=50x(10(2)^0.5/7) thus the ball travels 101m in the k vector

similary
for i vectors
d=20x(10(2)^0.5/7) thus the ball travels 40.41in the i vector

for the j vector, the ball lands on the ground so it has no height

hence we get final answer of
101i + 0j +40.41k +10i + 30k= 111i + 0i + 70.41K
i think..

I think I agree with most of the above UNTIL the last couple of lines. I get 50.4i +0j +131 k

Edit: Of course the above assumes j is vertical, I am beginning to think that k should be the vertical. To the OP is any mention made in the question about the vertical component.

Reply 17

Look at my 12.27 post - you've rediscovered the wheel the long way round!
We were there some time ago, except g was taken ass 10 which I reckon they wanted.

Reply 18