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M1 help!

For iii, why cant we we work out the time taken to reach the highest point (which is at zero velocity) then multiply it by 2 and then work out the horizontal distance by s=ut??
Please lemme know
please.PNG

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Reply 1
Avatar for Sinnoh
Sinnoh
22
Whilst you should find time of flight, your idea for finding it is wrong. It is not halfway through the time at its peak height, because it is projected from a point 15m above the ground.
You can find t from the fact that you have initial velocity 10 m/s, acceleration of -5 and displacement of -15. You don't need to break it up into two stages, that's just making your life difficult. The formula s=ut+12at2s = ut + \frac{1}{2}at^2 gets you two values of t, take the positive one. Then you do s=uts = ut
Original post by Kalabamboo
For iii, why cant we we work out the time taken to reach the highest point (which is at zero velocity) then multiply it by 2 and then work out the horizontal distance by s=ut??
Please lemme know


Twice the time to greatest height, takes you to the time when it's next as the same height as when it was projected.

In the case of A, starting from ground level, it will be the total time of flight.

In the case of B, starting from 15 m up, it will be the time when it's next 15m up, not the time when it hits the ground.
if it started and ended at the same level your method would work.
Reply 4
Avatar for Kalabamboo
Kalabamboo
OP
14
Original post by Sinnoh
Whilst you should find time of flight, your idea for finding it is wrong. It is not halfway through the time at its peak height, because it is projected from a point 15m above the ground.
You can find t from the fact that you have initial velocity 10 m/s, acceleration of -5 and displacement of -15. You don't need to break it up into two stages, that's just making your life difficult. The formula s=ut+12at2s = ut + \frac{1}{2}at^2 gets you two values of t, take the positive one. Then you do s=uts = ut

Thank you!
Reply 5
Avatar for Kalabamboo
Kalabamboo
OP
14
Original post by ghostwalker
Twice the time to greatest height, takes you to the time when it's next as the same height as when it was projected.

In the case of A, starting from ground level, it will be the total time of flight.

In the case of B, starting from 15 m up, it will be the time when it's next 15m up, not the time when it hits the ground.

Thank youuuuu!!! sorry doesn't lemme rep as you probs helped me quite recently:smile:
Reply 6
Avatar for Kalabamboo
Kalabamboo
OP
14
Original post by the bear
if it started and ended at the same level your method would work.


Thank youuuuu!!! sorry doesn't lemme rep as you probs helped me quite recently:smile:
Original post by Kalabamboo
Thank youuuuu!!! sorry doesn't lemme rep as you probs helped me quite recently:smile:


s'ok

man's got bare rep alredee

:gangster:
Reply 8
Avatar for Kalabamboo
Kalabamboo
OP
14
Original post by the bear
s'ok

man's got bare rep alredee

:gangster:

oh yhh:smile:thanks again:smile:
Reply 9
Avatar for Kalabamboo
Kalabamboo
OP
14
Guys.. how do you do part iv?:colondollar:
Reply 10
Avatar for Sinnoh
Sinnoh
22
Original post by Kalabamboo
Guys.. how do you do part iv?:colondollar:

If you make an equation trying to match the displacements of the particles with the same elapsed time, it should have no solutions if they do not collide.
s=ut+12at2s = ut + \frac{1}{2}at^2 is your friend.
(edited 5 years ago)
Original post by Sinnoh
Whilst you should find time of flight, your idea for finding it is wrong. It is not halfway through the time at its peak height, because it is projected from a point 15m above the ground.
You can find t from the fact that you have initial velocity 10 m/s, acceleration of -5 and displacement of -15. You don't need to break it up into two stages, that's just making your life difficult. The formula s=ut+12at2s = ut + \frac{1}{2}at^2 gets you two values of t, take the positive one. Then you do s=uts = ut


Original post by ghostwalker
Twice the time to greatest height, takes you to the time when it's next as the same height as when it was projected.

In the case of A, starting from ground level, it will be the total time of flight.

In the case of B, starting from 15 m up, it will be the time when it's next 15m up, not the time when it hits the ground.


Original post by the bear
if it started and ended at the same level your method would work.

Could you please help with part iv if you get the chance? Thanks a lot:smile:
Original post by Kalabamboo
Could you please help with part iv if you get the chance? Thanks a lot:smile:


Note the wording of the question, VERIFY.

So, work out the position of A, 0.75 seconds after A was projected.

And for B, 2.75 seconds after B was projected (as it was projected 2 sec earlier).

And show they represent the same position, horizontally and vertically.
Original post by Sinnoh
If you make an equation trying to match the displacements of the particles with the same elapsed time, it should have no solutions if they do not collide.
s=ut+12at2s = ut + \frac{1}{2}at^2 is your friend.

Im still sorta confused heres what the ms says
wwn.PNG
Original post by ghostwalker
Note the wording of the question, VERIFY.

So, work out the position of A, 0.75 seconds after A was projected.

And for B, 2.75 seconds after B was projected (as it was projected 2 sec earlier).

And show they represent the same position, horizontally and vertically.

Thanks but what do you do for IV ? Although I do think this will be highly useful for part v:smile:
Reply 15
Avatar for Sinnoh
Sinnoh
22
Original post by Kalabamboo
Im still sorta confused heres what the ms says
wwn.PNG


Okay, so there's actually barely any calculations necessary.
Because the total distance traveled by A and B are 60 and 70m respectively, their paths must be crossing because they are only 70m apart to begin with and they're travelling in opposite directions.
Then, because they both have the same initial vertical velocity and are projected at the same angle, their trajectories have similar shape.
Original post by Kalabamboo
Thanks but what do you do for IV ? Although I do think this will be highly useful for part v:smile:


Yep - should have put my other glasses on!

Sinnoh is giving you good advice on this one.
Reply 17
Avatar for Sinnoh
Sinnoh
22
For part v, substitute in (t−2)(t - 2) as the time travelled for A.
Original post by Sinnoh
Okay, so there's actually barely any calculations necessary.
Because the total distance traveled by A and B are 60 and 70m respectively, their paths must be crossing because they are only 70m apart to begin with and they're travelling in opposite directions.
Then, because they both have the same initial vertical velocity and are projected at the same angle, their trajectories have similar shape.

ah ok thanks and for the bit where they say " 0<t≤2 ) both have same vertical
motion so B is always 15 m above A " -why specifically between t=0 and t=2? and what do they mean by vertical motion? vertical velocity , vertical acceleration?
Thanks:smile:
Reply 19
Avatar for Sinnoh
Sinnoh
22
Original post by Kalabamboo
ah ok thanks and for the bit where they say " 0<t≤2 ) both have same vertical
motion so B is always 15 m above A " -why specifically between t=0 and t=2? and what do they mean by vertical motion? vertical velocity , vertical acceleration?
Thanks:smile:


Vertical motion just means in general, their speed and their acceleration, where they move. It's between 0s and 2s because after that, one has hit the ground.

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