physics

A gunman standing on a sloping ground fires up the slope. The initial speed of the bullet is v0= 390 m/s. The slope has an angle α= 19 degrees from the horizontal, and the gun points at an angle θ from the horizontal. The gravitational acceleration is g=10 m/s2.

(a) For what value of θ ( where θ>α) does the gun have a maximal range along the slope? (in degrees, from the horizontal)

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  1. (b) What is the maximal range of the gun, lmax, along the slope? (in meters)

    lmax=

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    posted by JuanPro
  2. Try to do 8:01 problems yourself first and post your attempts.
    These problems tend to take a long time, particularly for people who took the course 58 years ago. We are all volunteer teachers here and expect some effort by students.

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    posted by Damon
  3. lmax= (v_0^2*(1-sin(alpha))/(g*cos^2(alpha))

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  4. a)
    (pi/2)+(alpha/2)

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  5. I supposed that when bullet hits the slope, velocity on axis Y will be zero and it gives me one equation. I also have two equations, one for x axe(distance), and one for y axe(also distance, line perpendicular to the slope at x = cos Alpha * Lmax).
    After I solved it, I obtain an answer for theta which is less than 45 degrees and more that Alpha. Can it be a correct answer? I know that if there is no slope, just horizontal ground, max distance will be if shout at 45 degree.. So I doubt in my answer

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  6. The angle should be less than 45 degrees

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  7. the angle should be more than 45°!

    45° is for a horizontal plane.

    so theta = 45 + (alpha/2)

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  8. The angle should be less than 45 degrees

    how is that possible if 45+a/2 is the correct answer then the angle is more than 45 degrees..

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    posted by Greco
  9. the angle is (90+alpha)/(2)

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  10. Why 90? assuming alpha=30 (let's just say) then the resulting angle will be 60 which does not make sense since the plane is already inclined at 30 degrees.

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  11. Ok I see it has not to be 45. If the slope is let say 55 that your angle will need to be larger than 45 and minor than 90. I would go (90+55)/2.

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  12. because if you look closely the frame of reference is not a horizontal one. It's a little pink line and so by taking 90+alpha/2 you get the equivalent of 45 in that frame of reference

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  13. i got the green tick so the right answer is
    a) 45 + (alpha/2) (45 is the max angle without incline plus the a/2 for the incline

    b) (v_0^2*(1-sin(alpha))/(g*cos^2(alpha))

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    posted by Greco

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