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Thread: projectile fired

  1. #1
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    projectile fired

    a particle is projected from a point P on a horizontal plane with initial speed 35 m/s at an angle a to the plane. show that if x and y are the horizontal and vertical distances of the particle from P then 250y=250(tan a)x-(1-tan^2 a)x^2
    im not sure where to go with this but i was making the equation 35sin a t-4.9t^2=y and x=35cos a t dont no where to go from here
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  2. #2
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    Re: projectile fired

    You can rearrange your equation for x to express t as a function of x. Then substitute that back into your equation for y. This will give you an equation for y in terms of x. The only "trick" is that you'll end up with the second term being x^2/cos^2a, so you'll need to apply the identity 1+tan^2 a = 1/cos^2a. Note that the correct answer has a plus sign in front of the tan^2 term, not a negative sign.
    Last edited by ChipB; Dec 16th 2016 at 10:50 AM.
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  3. #3
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    Re: projectile fired

    when i sustitute t in i get xtana- (gx^2)/(2u^2cos^2a) how do you go from here to the answer and you were right about the +
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  4. #4
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    Re: projectile fired

    start with some trig identities ...

    $\dfrac{1}{\cos^2{a}} = \sec^2{a} = 1+\tan^2{a}$

    ... then note that $\dfrac{g}{2u^2} = \dfrac{1}{250}$ using your known/given values for $g$ and $u$.
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