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katykitz

  • 4 years ago

pls help... for a projectile launched from ground at an angle theta with the vertical , the maximum height reached is a. directly proportional to sin^2θ b. directly proportional to cos^2θ

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  1. arvind.satya
    • 4 years ago
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    a

  2. Niyor
    • 4 years ago
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    b. If the projectile is launched with an angle theta with the horizontal then only the answer is a. The formula to calculate height is: h= {(u^2)(sin^2θ)}/2g ; where u= initial velocity, g=acceleration due to gravity, θ = angle of projectile with the horizontal.

  3. shine
    • 4 years ago
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    a. directly proportional to sin^2θ

  4. ankurchiitd
    • 4 years ago
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    Lets see - if it was proportional to cos^2(theta) then a projectile launched vertically should have reached the least distance - which clearly is not true. Hence, sin^2(theta). It helps sometimes to put in 0 and 90 as the values of angles.

  5. ankitjain
    • 4 years ago
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    b, becs, the vertical component of the velocity would be V cos(θ), and by applying the eq. of motion, v^2-u^2=2(a)(h) where, v=0, u= V cos(θ) &a=(-g), we find that the height, h is directly proportional to cos^2(theta). @ankurchiitd: Cos(0)=1 is the max. value for the cosine function, and when the projectile would be launched vertically, i.e. with 0 degree to the vertical, it would attain the max. height.

  6. shadowfiend
    • 4 years ago
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    Note that incorrect answers are not considered abuse. If you see an incorrect answer, do as you did and simply correct it.

  7. katykitz
    • 4 years ago
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    umm...guys tnx 4 ur help...but could some1 pls tell me the exact answer...its so confusing !!!

  8. shadowfiend
    • 4 years ago
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    Several people posted the exact answer :)

  9. katykitz
    • 4 years ago
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    which is it?? ....a? pls tell me!

  10. shadowfiend
    • 4 years ago
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    Check out the answer that has two medals... Vs the ones that have none ;)

  11. katykitz
    • 4 years ago
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    k ...got it...tnx a lot.. :)

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