Find the correct launch angle

[Final_HB]

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.

Edited August 15, 2013 by Final_HB

[studiot]

OK, its your turn to put your thinking cap on and do some algebra/trigonometry

Edited August 15, 2013 by studiot

[Final_HB]

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 ?

[studiot]

I see you have moved on from comparing times in post#1 to distances in post#2.

 

However your formula are not correct.

 

D1 and D2 are not ranges.

 

Distance comparison is the correct approach, but you need to remember your formulae give you the range.

 

You have one further piece of information that is the range to C and its relation to D1 and D2.

 

So (R_a+ D1) = (R_b- D2) = R_c

Edited August 15, 2013 by studiot

[Final_HB]

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?

[studiot]

So what happens if you substitute this for R_a , R_band R_c and angles a,b and c in my equations above?

[Final_HB]

Doing that we get:

 

u²Sin2a +D1 = u²Sin2b - D2 =u²SinC

g g g

->u²(Sin2a-Sin2b) D2+D1=u²SinC

g g

[studiot]

Go on.

 

You have two equations connecting the quantities in your original expression plus one unknown (u) and wish to eliminate the one unknown between them.

[Final_HB]

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 :/

Edited August 16, 2013 by Final_HB

[studiot]

I'm glad you did the algebra/trig for yourself that is the idea of these forums.

 

You should now check the expression you originally posted in post#1.

 

Edited August 16, 2013 by studiot

[Final_HB]

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

[studiot]

Rest assured the expression with the additional pair of 2s is correct.

 

Both the missing 2's and the signs demonstrate the need for care in working.

 

If it helps, I get

 

D2 = V²/g(sin2b-sin2c)

 

D1 = V²/g(sin2c-sin2a)

 

divide the equations, cross multiply and collect terms.

[Final_HB]

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

 

Thanks for your help