Final_HB

ya looking back i never swapped signs after bring u2 across, so I missed it there Thanks for your help

i know... I do have it somewhere in my head... just I need someone to talk me through it sometimes thanks the original should be sin2a and sin2b not just sin a and sin b as i have :/ and the wrong signs are still an issue, but it might just be my algebra

Perfect... Just wanted to be sure I was on the right track with it before doing more work. Move some stuff around in both to get it equal to u2 so both equations become: u2=gsin2C-gsin2a and u2= gsin2C-gsin2b D1 D2 Let them equal each other. g cancels. remove fraction and you get: D1Sin2C-D1sin2b=D2sin2C-D2Sin2a And from there it pops out after some factorising EDIT:i actually ended with minuses in instead of pluses :/

Doing that we get: u2Sin2a +D1 = u2Sin2b - D2 =u2SinC g g g ->u2(Sin2a-Sin2b) D2+D1=u2SinC g g

Sorry... R= u2Sin2(theta) is the range formula the general one anyways. g Would you be able to say that the angle C would be between a and b?

Okay then Inital launch speed is u. Time of flight can be found using one of the main physics formulas, which comes out as: T=2uSina g g=acceleration due to gravity And our range for the first lunch comes out as: D1= 2u2Cos(a)sin(a) = 2u2Sin(2a) g g -The work is the exact same for D2, except at angle b so D2 = 2u2Sin2b g We could do this again, with angle C. Would the range be D2-D1 ?

When launched at an angle a to the horizontal, a projectile falls a distance D1 short of a target. If the projectile is fired at an angle b to the horiztonal, it reaches a distance D2 beyond the target. Assuming that C is the correct angle for a launched projectile to hit the target, show that: Sin2C = D2Sina + D1Sinb D1+D2 Well... If its a 'Sin2C' identity, it probably comes from Cosine times Sine. It probably goes like: -Find time of flight -Find a formula expressing the launch angle. -rinse and repeat for the second launch. -Equate the two, and mess with it til the desired result. EDIT: question states that both launch speeds are the same.

Cause the sum is 0, they can be just taken out of the formula and so F1+F2=0 is the only forces needed for equilibruim

Fair enough... The sum of the three forces is 0.

Sorry, I may have been unclear. I think I was referring to the same thing you are... But you used the correct terminology

Thank you Just making sure before I progress through it

I think so. something like: F1+F2+F3+F4+F5=0 And when its reversed: F1+F2-F3-F4-F5=0 let the two equal each other and cancel?

A particle is under the influence of 5 forces. Three of these forces are reversed, and the particle remains in equilibrium. Prove that the particle will remain in equilibrium even if these three forces were removed altogether. My thinking for this is: With three of the forces gone, the 2 forces (F1 and F2 ) left must balance the system if its to remain at equilibrium. If change of forces occurs, the system will react by moving to a new state of equilibrium. No reaction means that the 3 forces in the question cancel out each other, and no shift occurs. If the three forces balance to 0, there is no need to have them as acting on the particle. Right? If im right, Is there any maths way to say this, or a better way to explain what im trying to say. Thank you in advance.

A force of magnitude F acting up and along a smooth inclined plane, can support a mass M in equilibrium. If a force of the same magnitude acts horizontally, it can support a mass m on the same inclined plane in equilibrium. Find a relationship between F, M and m which is independent of the angle of inclination of the slope. Im not so sure on this one... I guess you: Take each force and particle separately. Resolve the forces into horizontal/vertical components. Let the sum of all these forces equal 0. Do the same for the second particle and equate the two equations to each other through F. Good so far?

Im glad you enjoy them I might have to spam a few more here