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Please can someone help me on question 9. The “half a second later” aspect is confusing me.

https://imgur.com/a/g8317nr

https://imgur.com/a/g8317nr

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#2

(Original post by

Please can someone help me on question 9. The “half a second later” aspect is confusing me.

https://imgur.com/a/g8317nr

**Y12_FurtherMaths**)Please can someone help me on question 9. The “half a second later” aspect is confusing me.

https://imgur.com/a/g8317nr

That should give you enough to find the initial velocity

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(Original post by

You’ve made a good start. You know how long after dropping the first ball the collision occurs. Try to use that to work out how long the second ball is travelling for. (Remember the half second delay between balls)

That should give you enough to find the initial velocity

**Snoozinghamster**)You’ve made a good start. You know how long after dropping the first ball the collision occurs. Try to use that to work out how long the second ball is travelling for. (Remember the half second delay between balls)

That should give you enough to find the initial velocity

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#4

**Y12_FurtherMaths**)

Please can someone help me on question 9. The “half a second later” aspect is confusing me.

https://imgur.com/a/g8317nr

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#5

(Original post by

Are you saying that the 2nd ball with be travelling for 1.75-0.5=1.25 seconds?

**Y12_FurtherMaths**)Are you saying that the 2nd ball with be travelling for 1.75-0.5=1.25 seconds?

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#6

**Y12_FurtherMaths**)

Please can someone help me on question 9. The “half a second later” aspect is confusing me.

https://imgur.com/a/g8317nr

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(Original post by

If you call the time 't' for the first then, in terms of t, what is the time fir the second?

**Muttley79**)If you call the time 't' for the first then, in terms of t, what is the time fir the second?

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(Original post by

Well the second ball is half a second behind, so its time variable would be .

**RDKGames**)Well the second ball is half a second behind, so its time variable would be .

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#9

**Y12_FurtherMaths**)

Please can someone help me on question 9. The “half a second later” aspect is confusing me.

https://imgur.com/a/g8317nr

**both**if you are taking upwards as the +ve direction.

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(Original post by

Also your acceleration needs to be -9.8 for

**RDKGames**)Also your acceleration needs to be -9.8 for

**both**if you are taking upwards as the +ve direction.
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#11

**RDKGames**)

Well the second ball is half a second behind, so its time variable would be .

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#12

(Original post by

I’m taking downwards as positive? And it wouldn’t make a difference would it, as it’s always 9.8 down and -9.8 up?

**Y12_FurtherMaths**)I’m taking downwards as positive? And it wouldn’t make a difference would it, as it’s always 9.8 down and -9.8 up?

**one**direction.

Once you choose whether that direction is +ve or not, that gravitational acceleration has its sign

*fixed*for

**ALL**bodies it affects,

__regardless__of the direction in which they are going.

If you are taking downwards as positive, then your second ball needs

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(Original post by

No, gravity always acts in

Once you choose whether that direction is +ve or not, that gravitational acceleration has its sign

If you are taking downwards as positive, then your second ball needs

**RDKGames**)No, gravity always acts in

**one**direction.Once you choose whether that direction is +ve or not, that gravitational acceleration has its sign

*fixed*for**ALL**bodies it affects,__regardless__of the direction in which they are going.If you are taking downwards as positive, then your second ball needs

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#14

(Original post by

If I’m taking downwards as positive then if a ball is dropped from a window it will accelerate down at 9.8ms^-2. And then if the ball is being projected up it will rise up with acceleration -9.8ms^-2. In really confused now

**Y12_FurtherMaths**)If I’m taking downwards as positive then if a ball is dropped from a window it will accelerate down at 9.8ms^-2. And then if the ball is being projected up it will rise up with acceleration -9.8ms^-2. In really confused now

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#15

**Y12_FurtherMaths**)

If I’m taking downwards as positive then if a ball is dropped from a window it will accelerate down at 9.8ms^-2. And then if the ball is being projected up it will rise up with acceleration -9.8ms^-2. In really confused now

Suppose you're on the ground and you take downwards as positive.

You throw a ball upwards with some velocity, and since this velocity acts upwards at the moment of your throw, it is a -ve value. This acceleration of 9.8m/s^2 is the rate at which the velocity of the ball will change as it goes up. Since velocity is initially -ve, every second 9.8m/s are added on top of it. It will eventually reach 0 velocity (hence reach the max point of the throw) and then the velocity will increase into the +ve's which means the direction of motion is now towards the ground (ie. downward)

As you'd expect.

Not sure where you got the idea from that acceleration is different for objects under the same gravitational field depending on the direction they are going.

The only case where it would be acceptable to say a=9.8 for first object and a=-9.8 for second object is if you decide all of a sudden to take UPWARDS as +ve after calculating your value from the first object... but just as it sounds, it's unnecessarily more complicated.

Last edited by RDKGames; 2 years ago

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(Original post by

Not quite.

Suppose you're on the ground and you take downwards as positive.

You throw a ball upwards with some velocity, and since this velocity acts upwards at the moment of your throw, it is a -ve value. This acceleration of 9.8m/s^2 is the rate at which the velocity of the ball will change as it goes up. Since velocity is initially -ve, every second 9.8m/s are added on top of it. It will eventually reach 0 velocity (hence reach the max point of the throw) and then the velocity will increase into the +ve's which means the direction of motion is now towards the ground (ie. downward)

As you'd expect.

Not sure where you got the idea from that acceleration is different for objects under the same gravitational field depending on the direction they are going.

The only case where it would be acceptable to say a=9.8 for first object and a=-9.8 for second object is if you decide all of a sudden to take UPWARDS as +ve after calculating your value from the first object... but just as it sounds, it's unnecessarily more complicated.

**RDKGames**)Not quite.

Suppose you're on the ground and you take downwards as positive.

You throw a ball upwards with some velocity, and since this velocity acts upwards at the moment of your throw, it is a -ve value. This acceleration of 9.8m/s^2 is the rate at which the velocity of the ball will change as it goes up. Since velocity is initially -ve, every second 9.8m/s are added on top of it. It will eventually reach 0 velocity (hence reach the max point of the throw) and then the velocity will increase into the +ve's which means the direction of motion is now towards the ground (ie. downward)

As you'd expect.

Not sure where you got the idea from that acceleration is different for objects under the same gravitational field depending on the direction they are going.

The only case where it would be acceptable to say a=9.8 for first object and a=-9.8 for second object is if you decide all of a sudden to take UPWARDS as +ve after calculating your value from the first object... but just as it sounds, it's unnecessarily more complicated.

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#17

**Y12_FurtherMaths**)

If I’m taking downwards as positive then if a ball is dropped from a window it will accelerate down at 9.8ms^-2. And then if the ball is being projected up it will rise up with acceleration -9.8ms^-2. In really confused now

For the first object, you took the top of the building as your

**reference**point and downwards as positive. This does imply that the object's displacement at time of collision is , and .

Then you switch focus to your second object. You scrap the point of reference you followed above and instead take the bottom of the building as your reference point, and now upwards as +ve as well. This, again, implies that at the point of collision, we have and .

This is of course not a problematic approach for this question and you can follow it all you want as long as you understand what's going on...

But it does prove to be problematic in the following reshuffling of the question:

"A ball is dropped from a window 30m above ground. Half a second later, another ball is projected vertically upwards with speed from the ground, vertically below the window.

**Find the time at which the two objects collide.**"

The only difference is that I don't tell you they collide 15m above the ground, and I give you an initial speed of the second object. This prevents you from using your approach because you cannot use 'consider first object, then consider the second object inidividually' tactic. You must instead consider them both here.

The approach is to set their displacements equal to eachother since that's when they collide, and this leaves you with an equation in .

Correct approach (taking downwards as +ve, for instance, with the bottom of the tower as the point of reference):

Incorrect approach (as you would do following the same style):

Last edited by RDKGames; 2 years ago

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#18

(Original post by

Is the answer 18ms^-1 (2sf) because the book says 17 but my teacher said the book is wrong

**Y12_FurtherMaths**)Is the answer 18ms^-1 (2sf) because the book says 17 but my teacher said the book is wrong

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(Original post by

After another glance at what you wrote, I noticed you actually took the 'unnecessarily more complicated' route which is where your misunderstanding must be stemming from.

For the first object, you took the top of the building as your

Then you switch focus to your second object. You scrap the point of reference you followed above and instead take the bottom of the building as your reference point, and now upwards as ve as well. This, again, implies that at the point of collision, we have and .

This is of course not a problematic approach for this question and you can follow it all you want as long as you understand what's going on...

But it does prove to be problematic in the following reshuffling of the question:

"A ball is dropped from a window 30m above ground. Half a second later, another ball is projected vertically upwards with speed from the ground, vertically below the window.

The only difference is that I don't tell you they collide 15m above the ground, and I give you an initial speed of the second object. This prevents you from using your approach because you cannot use 'consider first object, then consider the second object inidividually' tactic. You must instead consider them both here.

The approach is to set their displacements equal to eachother since that's when they collide, and this leaves you with an equation in .

Correct approach (taking downwards as ve, for instance, with the bottom of the tower as the point of reference):

Incorrect approach (as you would do following the same style):

**RDKGames**)After another glance at what you wrote, I noticed you actually took the 'unnecessarily more complicated' route which is where your misunderstanding must be stemming from.

For the first object, you took the top of the building as your

**reference**point and downwards as positive. This does imply that the object's displacement at time of collision is , and .Then you switch focus to your second object. You scrap the point of reference you followed above and instead take the bottom of the building as your reference point, and now upwards as ve as well. This, again, implies that at the point of collision, we have and .

This is of course not a problematic approach for this question and you can follow it all you want as long as you understand what's going on...

But it does prove to be problematic in the following reshuffling of the question:

"A ball is dropped from a window 30m above ground. Half a second later, another ball is projected vertically upwards with speed from the ground, vertically below the window.

**Find the time at which the two objects collide.**"The only difference is that I don't tell you they collide 15m above the ground, and I give you an initial speed of the second object. This prevents you from using your approach because you cannot use 'consider first object, then consider the second object inidividually' tactic. You must instead consider them both here.

The approach is to set their displacements equal to eachother since that's when they collide, and this leaves you with an equation in .

Correct approach (taking downwards as ve, for instance, with the bottom of the tower as the point of reference):

Incorrect approach (as you would do following the same style):

Last edited by Y1_UniMaths; 2 years ago

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#20

(Original post by

How do you know that they will collide when their displacements are equal and is that always the true?

**Y12_FurtherMaths**)How do you know that they will collide when their displacements are equal and is that always the true?

is always the condition in this consistent approach. If we measure displacement from different points, then TBH it just complicates things further and perhaps (?) makes the problem unsolvable.

means initial displacement.

(?) Unsure here because I can't be bothered to think about it much, and certainly this is going too beyond for you since you're new to this topic.

Last edited by RDKGames; 2 years ago

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