You are Here: Home >< Maths

# Where did i go wrong? watch

1. Hi guys can u have a look at my working for question 81 and let me know where i went wrong.

I made the assumption that the minimum angle would be such that the vertical component of velocity would equal zero at a height of 3m.

Apparently the answer is A but i got D.

2. (Original post by Shaanv)
Hi guys can u have a look at my working for question 81 and let me know where i went wrong.

I made the assumption that the minimum angle would be such that the vertical component of velocity would equal zero at a height of 3m.

Apparently the answer is A but i got D.

I’m not sure what you did, but I just used trig.
Tanx=O/A
Arctan(O/A) = X
Arctan (6/3) = X
Arctan(2) = X
3. (Original post by Shaanv)
Hi guys can u have a look at my working for question 81 and let me know where i went wrong.

I made the assumption that the minimum angle would be such that the vertical component of velocity would equal zero at a height of 3m.

Apparently the answer is A but i got D.
You worked out the angle for the minimum velocity.

If the ball, in the limit, went infinitely fast, it would travel in a straight line to the top of the wall. And that would be when the angle is a minimum.
4. (Original post by Daniel100499)
I’m not sure what you did, but I just used trig.
Tanx=O/A
Arctan(O/A) = X
Arctan (6/3) = X
Arctan(2) = X
What did i miss?
5. (Original post by Shaanv)
What did i miss?
It’s a triangle, the adjacent side is 6 units long, the opposite side is 3 units long.
I think you over complicated a simple question
6. (Original post by Daniel100499)
It’s a triangle, the adjacent side is 6 units long, the opposite side is 3 units long.
I think you over complicated a simple question
But tan(x)=opposite/adjacent. In ur original post it appears that u used tan(x)=adjacent/opposite. Unless u used the other angle in the traingle. If so why did u use that angle as opposed to the one i have marked as theta on my diagram?

Sorry for all the questions this is something simple but for some reason its just not clicking.

Guess its the sunday feeling.
7. this is the working out from the back of the book but as far as i can tell there is a mistake, due to this i calculate the answer as 45 degrees. Do u agree?
8. (Original post by Shaanv)
But tan(x)=opposite/adjacent. In ur original post it appears that u used tan(x)=adjacent/opposite. Unless u used the other angle in the traingle. If so why did u use that angle as opposed to the one i have marked as theta on my diagram?

Sorry for all the questions this is something simple but for some reason its just not clicking.

Guess its the sunday feeling.
Hmmmm yes sorry I made a mistake it should just be arctan (3/6)
I even just put it into a triangle calculator online and it gave me an angle of arctan(1/2) so I don’t know why the answer is supposed to be A
9. (Original post by Daniel100499)
Hmmmm yes sorry I made a mistake it should just be arctan (3/6)
I even just put it into a triangle calculator online and it gave me an angle of arctan(1/2) so I don’t know why the answer is supposed to be A
(Original post by ghostwalker)
You worked out the angle for the minimum velocity.

If the ball, in the limit, went infinitely fast, it would travel in a straight line to the top of the wall. And that would be when the angle is a minimum.
Is there working right. I think theres a mistake which propagates through to give an answer of 45 degrees
Attached Images

10. (Original post by Shaanv)
Is there working right. I think theres a mistake which propagates through to give an answer of 45 degrees
Ohhh I see what you mean now, sorry it’s been 2 years since I did physics. The working in the book does seem wrong as you shouldn’t get 1/6 if you rearrange the equation you should get 1/12 which if you slot that back in you end up with arctan(1) which is 45 so yes I think the book is wrong
11. (Original post by Shaanv)
Is there working right. I think theres a mistake which propagates through to give an answer of 45 degrees
From the wording of the first sentence in the question, I get the impression it's been changed prior to publication. I go with arctan(1/2) as the minimum angle - though it's not actually achievable.

Question is a bit of a mess - I'd ignore it and move on.
12. (Original post by ghostwalker)
From the wording of the first sentence in the question, I get the impression it's been changed prior to publication. I go with arctan(1/2) as the minimum angle - though it's not actually achievable.

Question is a bit of a mess - I'd ignore it and move on.
Will do thanks a bunch
13. Thanks for this thread. I did it the same way as the OP and got D (hypo.) but I haven't learnt Trig yet.

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: October 29, 2017
Today on TSR

### University open days

• University of East Anglia (UEA)
Could you inspire the next generation? Find out more about becoming a Secondary teacher with UEA… Postgraduate
Thu, 18 Oct '18
• University of Warwick
Sat, 20 Oct '18
• University of Sheffield
Sat, 20 Oct '18
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