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Projectile fired along an incline

  • Thread starter asap9993
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  • #1
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Homework Statement


Some members here may have seen this problem before. The problem is:
A projectile is fired up an incline (incline angle φ) with an initial speed, v ,at an angle θ with respect to the horizontal (θ>φ). What is the angle that will allow the projectile to travel the maximum distance, r , along the incline plane and what is that value of r?

Homework Equations



r(θ) = [2(v^2)cos(θ)sin(θ-φ)]/g(cos^2(φ))

The Attempt at a Solution


When I differentiate r(θ) and set it equal to zero, I get the equation cot(2θ) = -tan(φ) which then becomes cot(2θ) = -cot(pi/2 - φ) because of the identity
cot(pi/2 - x) = tan(x).
Thus, 2θ = -(pi/2 - φ) and θ = φ/2 - pi/4

Please help!!

However the book says the answer is θ = φ/2 + pi/4
Did I make an error?
 

Answers and Replies

  • #2
tiny-tim
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welcome to pf!

hi asap9993! welcome to pf! :smile:

(have a pi: π and try using the X2 icon just above the Reply box :wink:)
r(θ) = [2(v^2)cos(θ)sin(θ-φ)]/g(cos^2(φ))

When I differentiate r(θ) and set it equal to zero, I get the equation cot(2θ) = -tan(φ)
how did you get that? :confused:

i get a coscos - sinsin formula

try again! :smile:
 
  • #3
19
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To tiny-tim,

After using the product rule, I got
dr/dθ = [(2v^2)/g(cosφ)^2][cos(θ)cos(θ-φ) - sin(θ)sin(θ-φ)] = 0 then

cos(θ)cos(θ-φ) = sin(θ)sin(θ-φ)

cos(θ)[cos(θ)cos(φ) + sin(θ)sin(φ)] = sin(θ)[sin(θ)cos(φ) - cos(θ)sin(φ)]

cos(φ)[(cosθ)^2] + cos(θ)sin(θ)sin(φ) = cos(φ)[(sinθ)^2] - cos(θ)sin(θ)sin(φ)

2cos(θ)sin(θ)sin(φ) = cos(φ)[(sinθ)^2 - (cosθ)^2]

Multiplying both sides of the above equation by -1 and using the two trig identities:
(cosθ)^2 - (sinθ)^2 = cos(2θ) and 2cos(θ)sin(θ) = sin(2θ) gives

-sin(2θ)sin(φ) = cos(2θ)cos(φ) and then

cot(2θ) = -tan(φ)
 
  • #4
tiny-tim
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hi asap9993! :smile:

i'll check the rest of it in a moment, to see where you've gone wrong

but you should have stopped at
dr/dθ = [(2v^2)/g(cosφ)^2][cos(θ)cos(θ-φ) - sin(θ)sin(θ-φ)] = 0
because the second bracket is already cosAcosB - sinAsinB,

sooo you can immediately say that it's … ? :wink:
 
  • #5
19
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OH SNAP!

Thank you so much tiny-tim. I can't believe I didn't see that! I guess when you're working with trig, you gotta know when to stop.
 
  • #6
tiny-tim
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hi asap9993! :smile:

i almost forgot …
When I differentiate r(θ) and set it equal to zero, I get the equation cot(2θ) = -tan(φ) which then becomes cot(2θ) = -cot(pi/2 - φ) because of the identity
cot(pi/2 - x) = tan(x).
Thus, 2θ = -(pi/2 - φ) and θ = φ/2 - pi/4
there's no error here, apart from the fact that cotA = cot B implies either A = B or A = B + π …

the second one gives you 2θ = π - (π/2 - φ) = π/2 + φ as required …

what is the other solution, then? it's for the minimum value of r, ie the farthest downhill if you launch the projectile at an angle less than the slope, φ :biggrin:
 

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