x Turn on thread page Beta
 You are Here: Home >< Maths

# Projectiles watch

1. Can anyone give me a few hints for 12b. and 6?
Attached Images

2. (Original post by pippabethan)
Can anyone give me a few hints for 12b. and 6?
For Q6, resolve the velocity horizontally and vertically.

Horizontally: (x is horizontal displacement)

Vertically: (y is vertical displacement)

Path of the projectiles means an equation of the form , so you want to eliminate from the second equation.

You know that , so substitute that into your second equation and that'll give you the path (which will be a quadratic function of , plotting it will look familiar to you, it'll be a maximum quadratic that models the path of the projectile quite literally).
3. (Original post by pippabethan)
Can anyone give me a few hints for 12b. and 6?
12 looks badass. I haven't seen this type before. For part 12a I think the idea is gradient=acceleration =g. But how did you mathematically show it? Mind sharing?
For part 12B the idea is that they have different constant horizontal speeds I.e. rate of change of distance with time is different for each. I would like to see the solutions if you do them!
4. (Original post by thebrahmabull)
12 looks badass. I haven't seen this type before. For part 12a I think the idea is gradient=acceleration =g. But how did you mathematically show it? Mind sharing?
For part 12B the idea is that they have different constant horizontal speeds I.e. rate of change of distance with time is different for each. I would like to see the solutions if you do them!
I'll post up the solutions if the OP doesn't.
5. (Original post by Zacken)
I'll post up the solutions if the OP doesn't.
Thanks
6. (Original post by Zacken)
I'll post up the solutions if the OP doesn't.
we are waiting ...
7. (Original post by TeeEm)
we are waiting ...
As I am for the OP to reply...
8. (Original post by Zacken)
As I am for the OP to reply...
sorry as I misunderstood ... just woke up
I thought you were putting them up.
9. (Original post by pippabethan)
Can anyone give me a few hints for 12b. and 6?
Oh well, since I've done Q12 - might as well put this up, since you're not replying:

So

The gradient of this line should be fairly easy to find, and it will be in terms of only and , which are constants.
10. (Original post by Zacken)
For Q6, resolve the velocity horizontally and vertically.

Horizontally: (x is horizontal displacement)

Vertically: (y is vertical displacement)

Path of the projectiles means an equation of the form , so you want to eliminate from the second equation.

You know that , so substitute that into your second equation and that'll give you the path (which will be a quadratic function of , plotting it will look familiar to you, it'll be a maximum quadratic that models the path of the projectile quite literally).
Thank you!
11. Sorry everyone, my wifi stopped working all day :/ Thank you so much for your help!! I can post the answers now thebrahmabull if you still want them?
12. (Original post by Zacken)
Oh well, since I've done Q12 - might as well put this up, since you're not replying:

So

The gradient of this line should be fairly easy to find, and it will be in terms of only and , which are constants.
I got the constant gradient part, but how do you find the rate of increase of distance? I think the formula v^2=u^2 + 2as is used because the answer involves a root.
13. (Original post by pippabethan)
I got the constant gradient part, but how do you find the rate of increase of distance? I think the formula v^2=u^2 + 2as is used because the answer involves a root.
Increasing at a constant rate means that the gradient of the distance separating them is positive. Can you prove that your answer to part (a) is positive?
14. (Original post by Zacken)
Increasing at a constant rate means that the gradient of the distance separating them is positive. Can you prove that your answer to part (a) is positive?
Of course! Thank you, I'll give it a go tomorrow
15. (Original post by Zacken)
For Q6, resolve the velocity horizontally and vertically.

Horizontally: (x is horizontal displacement)

Vertically: (y is vertical displacement)

Path of the projectiles means an equation of the form , so you want to eliminate from the second equation.

You know that , so substitute that into your second equation and that'll give you the path (which will be a quadratic function of , plotting it will look familiar to you, it'll be a maximum quadratic that models the path of the projectile quite literally).
Wouldn't t=x/20 ? I got 400y = 600x -4.9x^2 as the final answer, but the book says it's 900y
16. (Original post by pippabethan)
Wouldn't t=x/20 ? I got 400y = 600x -4.9x^2 as the final answer, but the book says it's 900y
Apologies, memory loss in the short time-span between reading the question from your picture and typing my stuff out, it's been fixed. The direction would be horizontal, so the initial horizontal speed is , so it would be .
17. I don't see how the rate of increase should equal u(2(1-cos(theta - alpha))^0.5.
For a) I got (sin theta - sin alpha)/(cos theta - cos alpha)
18. (Original post by Zacken)
Increasing at a constant rate means that the gradient of the distance separating them is positive. Can you prove that your answer to part (a) is positive?
I couldnt.
19. (Original post by TeeEm)
we are waiting ...
Lol

TSR Support Team

We have a brilliant team of more than 60 Support Team members looking after discussions on The Student Room, helping to make it a fun, safe and useful place to hang out.

This forum is supported by:
Updated: March 13, 2016
Today on TSR

### Complete university guide 2019 rankings

Find out the top ten here

Poll
Useful resources

### Maths Forum posting guidelines

Not sure where to post? Read the updated guidelines here

### How to use LaTex

Writing equations the easy way

### Study habits of A* students

Top tips from students who have already aced their exams

## Groups associated with this forum:

View associated groups

The Student Room, Get Revising and Marked by Teachers are trading names of The Student Room Group Ltd.

Register Number: 04666380 (England and Wales), VAT No. 806 8067 22 Registered Office: International House, Queens Road, Brighton, BN1 3XE