# M1 help!

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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

Please lemme know

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#2

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 gets you two values of t, take the positive one. Then you do

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 gets you two values of t, take the positive one. Then you do

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#3

(Original post by

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

**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

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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(Original post by

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 gets you two values of t, take the positive one. Then you do

**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 gets you two values of t, take the positive one. Then you do

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(Original post by

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.

**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.

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(Original post by

if it started and ended at the same level your method would work.

**the bear**)if it started and ended at the same level your method would work.

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#8

(Original post by

Thank youuuuu!!! sorry doesn't lemme rep as you probs helped me quite recently

**Kalabamboo**)Thank youuuuu!!! sorry doesn't lemme rep as you probs helped me quite recently

man's got bare rep alredee

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#11

(Original post by

Guys.. how do you do part iv?

**Kalabamboo**)Guys.. how do you do part iv?

is your friend.

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**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 gets you two values of t, take the positive one. Then you do

**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

if it started and ended at the same level your method would work.

**the bear**)if it started and ended at the same level your method would work.

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#13

(Original post by

Could you please help with part iv if you get the chance? Thanks a lot

**Kalabamboo**)Could you please help with part iv if you get the chance? Thanks a lot

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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(Original post by

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.

is your friend.

**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.

is your friend.

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(Original post by

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.

**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.

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#16

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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#17

(Original post by

Thanks but what do you do for IV ? Although I do think this will be highly useful for part v

**Kalabamboo**)Thanks but what do you do for IV ? Although I do think this will be highly useful for part v

Sinnoh is giving you good advice on this one.

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(Original post by

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.

**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.

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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#20

(Original post by

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

**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

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