Kalabamboo
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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
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Sinnoh
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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 + \frac{1}{2}at^2 gets you two values of t, take the positive one. Then you do s = ut
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ghostwalker
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(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.
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the bear
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if it started and ended at the same level your method would work.
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Kalabamboo
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(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 + \frac{1}{2}at^2 gets you two values of t, take the positive one. Then you do s = ut
Thank you!
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Kalabamboo
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(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
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Kalabamboo
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(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
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the bear
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(Original post by Kalabamboo)
Thank youuuuu!!! sorry doesn't lemme rep as you probs helped me quite recently
s'ok

man's got bare rep alredee

:gangster:
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Kalabamboo
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(Original post by the bear)
s'ok

man's got bare rep alredee

:gangster:
oh yhhthanks again
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Kalabamboo
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Guys.. how do you do part iv?
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Sinnoh
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(Original post by Kalabamboo)
Guys.. how do you do part iv?
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 + \frac{1}{2}at^2 is your friend.
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Kalabamboo
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(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 + \frac{1}{2}at^2 gets you two values of t, take the positive one. Then you do s = 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
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ghostwalker
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(Original post by Kalabamboo)
Could you please help with part iv if you get the chance? Thanks a lot
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.
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Kalabamboo
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(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 + \frac{1}{2}at^2 is your friend.
Im still sorta confused heres what the ms says
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Kalabamboo
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(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
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Sinnoh
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(Original post by Kalabamboo)
Im still sorta confused heres what the ms says
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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.
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ghostwalker
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(Original post by Kalabamboo)
Thanks but what do you do for IV ? Although I do think this will be highly useful for part v
Yep - should have put my other glasses on!

Sinnoh is giving you good advice on this one.
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Sinnoh
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For part v, substitute in (t - 2) as the time travelled for A.
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Kalabamboo
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(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
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Sinnoh
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(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
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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