3U Maths Question Thread

[Rabi]

Hey everyone I needed help with projectile question I tired them but couldn't get an answer to this question can you plz help. 11. The heights point of the trajectory of a body is 300m and the horizontal range is 5000m. Find the initial velocity and angle of projection.
12. A projectile reaches a max height of 100m above the point of projection at a horizontal distance of 250m from this point. Show that it was fired at an angle of elevation of approx 38*40* and calculate the max range for the same velocity of projection.

[jamonwindeyer]

Hey everyone I needed help with projectile question I tired them but couldn't get an answer to this question can you plz help. 11. The heights point of the trajectory of a body is 300m and the horizontal range is 5000m. Find the initial velocity and angle of projection.
12. A projectile reaches a max height of 100m above the point of projection at a horizontal distance of 250m from this point. Show that it was fired at an angle of elevation of approx 38*40* and calculate the max range for the same velocity of projection.

Hey! The principle for both is pretty much the same, since the range is given in 11, and half the range is given in 12. So, at the peak of motion, vertical velocity is zero:



But this occurs at a height of 300 metres, so we can now go to displacement:



That's one equation. We also know that the horizontal range is 5000m, and this occurs at double the time of the maximum height. So we can say:



Substitute that first equation into the second:



From there you can solve for theta (the acute angle) and use that to find the velocity! The working is almost identical for 12! There might be a faster way, but this is how I first thought to approach it (also did this quick so there might be little numerical errors, I'm doing the numbers in my head)

[Rabi]

Ohhh ok cool tnx

[Rabi]

Can someone please help and show me how to do this question I am not getting the answer to question 9 and 10. I don't know how to find velocity. Pls help.

[jamonwindeyer]

Can someone please help and show me how to do this question I am not getting the answer to question 9 and 10. I don't know how to find velocity. Pls help.

Hey! The method for Q9 is exactly the same as I showed you above, try and copy it over!

As for Q10, a little different, but similar. At the highest point, the vertical velocity is zero, which means that the horizontal velocity is equal to 10 metres per second at that point. However, horizontal velocity is constant, so this immediately gives us a formula linking velocity and the angle



We can also use the formula we derived above, obtained by setting the vertical velocity to zero;



Now we also know it is 8 metres above the ground at this point, so (again, similar to above):



Sub that back in to find the angle!

[spnmox]

Hi, I'm having trouble with this inequality question:
Prove that, if a,b,c and d are any four positive numbers, then ab+cd <= sqr((a^2+c^2)(b^2+d^2))
 I worked backwards and got that (a^2+c^2)(b^2+d^2) >=4abcd, and I need to prove that (ab+cd)^2<=4abcd. I don't know how to manipulate the LHS to do that, and also is this the right method?

Also, just generally: what's the best way to approach these kinds of inequality questions?
What are some useful inequality identities to know eg. a^2+b^2>=2ab or are you expected to derive them from the beginning?

I really appreciate you answering my questions!

[RuiAce]

Hi, I'm having trouble with this inequality question:
Prove that, if a,b,c and d are any four positive numbers, then ab+cd <= sqr((a^2+c^2)(b^2+d^2))
 I worked backwards and got that (a^2+c^2)(b^2+d^2) >=4abcd, and I need to prove that (ab+cd)^2<=4abcd. I don't know how to manipulate the LHS to do that, and also is this the right method?

Also, just generally: what's the best way to approach these kinds of inequality questions?
What are some useful inequality identities to know eg. a^2+b^2>=2ab or are you expected to derive them from the beginning?

I really appreciate you answering my questions!

These questions are not a part of the 3U course; they form the harder 3U section of the 4U syllabus. Please post them there if you require assistance.

Jamon Edit: Here is the link!

[spnmox]

These questions are not a part of the 3U course; they form the harder 3U section of the 4U syllabus. Please post them there if you require assistance.

Jamon Edit: Here is the link!

Oh, oops. Thanks for the clarification!

[Rabi]

Thank you so much but I have used the same process but I am not getting the answer for both questions!!!!!

[jamonwindeyer]

Thank you so much but I have used the same process but I am not getting the answer for both questions!!!!!

As in you are getting the incorrect answer? Or were you having trouble applying the method?

If incorrect answer, pop a picture of your working up if you can! We'll probably be able to spot the issue

If having trouble, let us know whereabouts you get stuck!

[annabeljxde]

Hey everyone. The last couple of questions for 3U trig in my textbook is getting really challenging and I was hoping if I could get help on these questions.

There's quite a lot so I'll separate them into 3 different posts. (I hope its not too much,, :') )

Thank you so much!


QUESTION 7: I can't seem to get the correct answer. The answer is 1/24(16-9√3)

[annabeljxde]

QUESTION 9e)

I tried using a u-substitution method that Symolab used but I still didn't get the correct answer.

Answer is 1/4(1/3sin3θ + 3sinθ) + C

[annabeljxde]

QUESTION 10b)

I squared the equation I got in part a) but I didn't get the right answer...

Answer is π/4(2-√3)units^3

[jamonwindeyer]

Hey everyone. The last couple of questions for 3U trig in my textbook is getting really challenging and I was hoping if I could get help on these questions.

There's quite a lot so I'll separate them into 3 different posts. (I hope its not too much,, :') )

Thank you so much!


QUESTION 7: I can't seem to get the correct answer. The answer is 1/24(16-9√3)

I'll take this and the next one! The expression from (a) is:



So your integral becomes:



Substitute and solve!

For your next one, your answer is correct but written in a different form I believe! If you substitute the expression I have for \(\sin^3{x}\) from above, then you should get a match with theirs!

[fun_jirachi]

QUESTION 10b)

I squared the equation I got in part a) but I didn't get the right answer...

Answer is π/4(2-√3)units^3

Jamon took the above two I believe? So I'll take this one

When you're dealing with volumes, in your first line I can already see you've made a mistake loads of people make. The key thing here is that you need to take top curve squared minus bottom curve squared, not top curve minus bottom curve all squared.



What you've done is great, you've just started with the wrong integral!

Hope this helps