M2 problem involving motion

stuck on the first part of question 2
i have narrowed it down to

u^2 > \frac{ag}{2cos^2\theta(tan\alpha +tan\theta)}

i tried subbing values in my equation and the same values in the textbook answer, and they gave the same answer,does anyone know how i convert my equation into the equation on question 2? or have i gone wrong?

(edited 10 years ago)

Reply 1

ghostwalker

Original post by physics4ever

u^2 > \frac{ag}{2cos^2\theta(tan\alpha +tan\theta)}

Your equation is correct so far.

I'd start by multiplying the top and bottom by

\cos\alpha

to get the numerator you want.

Then consider what you're aiming for and what the expansion of

\sin(2\theta+\alpha)

looks like and try and aim for it.

Reply 2

physics4ever

Original post by ghostwalker

Your equation is correct so far.

I'd start by multiplying the top and bottom by

\cos\alpha

to get the numerator you want.

Then consider what you're aiming for and what the expansion of

\sin(2\theta+\alpha)

looks like and try and aim for it.

okay thanks for the help! :smile:

Reply 3

physics4ever

Original post by ghostwalker

Your equation is correct so far.

I'd start by multiplying the top and bottom by

\cos\alpha

to get the numerator you want.

Then consider what you're aiming for and what the expansion of

\sin(2\theta+\alpha)

looks like and try and aim for it.

just tried it and i couldnt do it

EDIT: actually i spotted that i havent got it, i ended up with

u^2>\frac{agcos(\alpha)}{2cos( \theta )sin(\theta+\alpha)}

(edited 10 years ago)

Reply 4

physics4ever

actually i spotted that i havent got it, i ended up with

u^2>\frac{agcos(\alpha)}{2cos( \theta )sin(\theta+\alpha)}

(edited 10 years ago)

Reply 5

ghostwalker

Original post by physics4ever

actually i spotted that i havent got it, i ended up with

u^2>\frac{agcos(\alpha)}{2cos( \theta )sin(\theta+\alpha)}

Yes, one arrangement of the terms leads you to that.

You want to focus on the expansion of sin(2theta + alpha), and getting things into that form.

Post some working if you wish.

Reply 6

physics4ever

Original post by ghostwalker

Yes, one arrangement of the terms leads you to that.

You want to focus on the expansion of sin(2theta + alpha), and getting things into that form.

Post some working if you wish.

just got the answer, thats a lot of rearanging for a question 2

Reply 7

BabyMaths

Original post by physics4ever

actually i spotted that i havent got it, i ended up with

u^2>\frac{agcos(\alpha)}{2cos( \theta )sin(\theta+\alpha)}

I see you have it now but I thought I'd mention that the denominator you have here can be changed to the required form by using the product and sum formulae.

Reply 8

physics4ever

Original post by BabyMaths

I see you have it now but I thought I'd mention that the denominator you have here can be changed to the required form by using the product and sum formulae.

i was actually trying to use the addition formulae,but it never occured to me that they can be rearanged like that, thanks for showing me that! :smile: