Projectile Motion (1 Viewer)

[CrashOveride]

G'day :uhhuh:

Question: A body is projected with speed U from a height h, above a horizontal plane, at an angle &thetha; to the horizontal. Show that the range R on the horizontal is given by:

gR²sec²θ - 2U²Rtanθ - 2hU² = 0.

Hokay, did this first part then the next part says: Further show that the maxium range R₁ is given by:

R₁ = (U/g) . sqrt[ U² + 2hg ]

So my approach was to solve the quadratic in R from above, keeping in mind that we are after a maximum answer. Except i wasn't too sure were to go from here, i have attached an image (easier to read than writing here i take it) where i think im close just anyone got any hints from here? Thanks !

 

[ngai]

try to eliminate theta from your equations, and deal with time t instead

 

[CrashOveride]

Well i was thinking i need to get it so i have a maxium (obviously). Except which angle could i say that... because the left hand side it would differ from the RHS.

 

[ngai]

x = utcos@
y = utsin@ - 1/2gt^2 + h
and u want max x, when y = 0
so utcos@ = x
y = 0, so utsin@ = 1/2gt^2 - h
x^2 + (1/2gt^2 - h)^2 = (ut)^2
now can u find maximum x?

 

[CrashOveride]

Ok ive tried like 4 different ways now that all seem to lead me astray.

Going from your last suggestion, i went dx/dt = 0 and got time to max range then put that in to x=utcos@ but it doesnt seem to work.

 

[wogboy]

g*(R^2)*sec^2(@) - 2*(U^2)*R*tan@ - 2*h*U^2 = 0

so by implicit differentiation:
dR/d@ = -{2g*(R^2)*sec^2(@)*tan@ - 2*(U^2)*R*sec^2(@)}/{2*g*R*sec^2(@) - 2*(U^2)*tan@}

setting dR/d@ = 0,
2g*(R^2)*sec^2(@)*tan@ = 2*(U^2)*R*sec^2(@)
g*R*tan@ = (U^2)
tan@ = U^2 /(g*R)
sec^2(@) = tan^2(@) + 1
= U^4/(g^2*R^2) + 1

subbing tan@ and sec^2(@) back into the original equation:
g*(R^2)*U^4 / (g^2*R^2) + g*R^2 - 2*(U^2)*R*(U^2)/(g*R) - 2*h*U^2 = 0
(U^4)/g + g*R^2 - 2*(U^4)/g - 2*h*U^2 = 0
g*R^2 = 2*h*U^2 + (U^4)/g
R^2 = 2*h*(U^2)/g + (U^4)/(g^2)
= (U/g)^2 * (2gh + U^2)
hence R = U/g * sqrt(2gh + U^2)

 

[CrashOveride]

Thanks wogboy!

Just curious, what if we took say d@/dR. Should it give the same result? I tried and i get to a point where i can first solve for R, but it doesnt give me the required max range. Whats the reasioning behind that and whats the reasoning behind say targetting d@/dR instead of dR/d@ or even something like dR/dU etc etc. ?

Thanx !

 

[wogboy]

It's all a matter of identifying the roles of each variable. R is the dependent variable (which you wish to maximise), @ is the independent variable (which you can arbitrarily vary), and U, h etc are all controlled variables (fixed values, invariant).

If I was to give you a simple function y = f(x) to plot, and asked you to find the maxima/minima of the function, you would set dy/dx = 0 (not dx/dy = 0). y is the dependent variable which you wish to maximise/minimise, and x is the dependent variable which you can vary arbitrarily. Anything else will be a controlled variable (i.e. constants)

So in this case you wish to maximise R and the variable which you are allowed to vary is @, so you set dR/d@ = 0, no other way accepted.

 

[CrashOveride]

Makes sense.

A particle is projected from a point O with velocity v at an angle θ to the horizontal. It passes through the point P(x,y) in the vertical plane through O where (x,y) are the co-ords of P w.r.t. rectangular axes at O.

Prove that y = xtan@ - (gx²sec²@)/ 2v². <B>Done</B>

If x = 20m, y = 10m, g = 10, v = 20 find the two values of tan@ [<B>Done</B>] and using t=tan@ and sin2@ = 2t / (1 + t²) , prove that the ratio of two ranges is 5/3.

Hmm well for thw two values of tan@ its tan@ = 3 or 1. I tried doing it just substituting this into the range equation (setting that thing up there y=0). But i get like that the range of the smaller larger angle is 1/6 than that of the 45degree one. Im not sure how to incorporate sin2@ here ?

 

[wogboy]

If x = 20m, y = 10m, g = 10, v = 20 find the two values of tan@ [Done] and using t=tan@ and sin2@ = 2t / (1 + t2) , prove that the ratio of two ranges is 5/3.

setting y = 0 (for the range)

x*tan@ - (1/80)*(x^2)*sec^2(@) = 0
80*tan@ - x*sec^2(@) = 0 (since the range, x != 0)
x = 80*tan@*cos^2(@)
= 80*sin@*cos@
= 40*sin(2@)

You found that t = tan@ = 3 or 1,
so sin(2@) = 2t/(1 + t^2)
= 0.6 or 1

the ratio of the two ranges
= x2 / x1
= (40*1) / (40*0.6)
= 1/0.6
= 5/3

 

[BlackJack]

Hmm... looks done, crash. You didn't need me.

 

[CrashOveride]

Thanks wogboy.

Dont worry BJ, since you are so eager, ill make sure to get a question for YOU soon

EDIT: I actually had the range in another form and all i had to do was sub in the tan result from above, doh. no need for this sin2@

 

[CrashOveride]

Say we had the same flight path as above y = xtan@ - (gx^2.sec^2(@))/ 2v^2

For a given range R there are two possible directions pf projection tan@₁ and tan@₂. Let T₁ and T₂ be the times of flights correspdoning to angles of projections. I proved T1/T2 = sin@1 / sin@2 and also that R = R₁.sin2@ (where R1 is max range)

Now if T1 / T2 = 2, show R / R₁ = 4/5 And its given the hint to use sin2@ in terms of 't' results

 

[wogboy]

Now if T1 / T2 = 2, show R / R1 = 4/5 And its given the hint to use sin2@ in terms of 't' results

T1/T2 = 2
-> sin(@1) / sin(@2) = 2
-> sin(@1) = 2*sin(@2)

R/R1 = sin(2*@1) = sin(2*@2)
-> 2@1 = pi - 2@2 (since sin2@ is symmetric about @ = pi, and we know @1 & @2 are between 0 and pi/2)
-> @2 = pi/2 - @1

from above, sin(@1) = 2*sin(@2)
-> sin(@1) = 2*sin(pi/2 - @1)
-> sin(@1) = 2*cos(@1)
-> tan(@1) = 2
-> sin(2*@1) = 2*tan(@1) / (1 + tan^2(@1))
= 2*2 / (1 + 4)
= 4/5
-> R/R1 = 4/5

 

[redslert]

hehe you told me to take a look
and all i can say is that i have forgotten all my 4u aproximately 30mins after the 4u exam!!

 

[CrashOveride]

i was intersted in your solutiosn u had as u said u had totally completed the patel 4unit book ?

 

[redslert]

hehe all my 4u work along with everything from highschool is tucked away in a nice little box somewhere which i don't feel like looking at for another couple of decades!

although i gave away most of my maths stuff, i still have my exercise books...but looking at wogboy's solutions, it pretty much is the same