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Thread: projectile

  1. #1
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    projectile

    A particle is projected from a point on the horizontal floor of a tunnel with maximum height of 8 m.
    The particle is projected with an initial speed of u m/s.
    show the greatest range which can be attained in the tunnel is 8√(u^2-16g/g)

    i have gotten the time equal to 2usina/g i have also worked out sina=4√g/u dont know how to go any further
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  2. #2
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    Re: projectile

    I don't think this is necessarily correct.

    They found this answer by assuming the highest point of the trajectory above the floor will be 8m.

    That's fine as long as the initial velocity is fast enough that a firing angle of less than or equal to 45 degrees will accomplish this.

    If the velocity is slower than this then the optimal firing angle for range will be 45 degrees and the formula given is incorrect.

    To obtain the formula they give just note

    i) the high point of the trajectory is found where $v_y=0$ Call this time $\dfrac{t_f}{2}$

    ii) at $t=\dfrac{t_f}{2},~y=8$

    ii) the range is $u \cos(\theta) t_f$

    you can first calculate the midpoint time and then the firing angle. This will give you the range in terms of the initial velocity.
    Last edited by romsek; Jan 28th 2017 at 09:09 AM.
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    Re: projectile

    How did you get the time? The time at which what happens?

    If the particle is projected with speed u at angle a then at time t the height above the floor is h= (-g/2)t^2+ u sin(a) t. The maximum height occurs when h'= -gt+ u sin(a)= 0 of t= u sin(a)/g

    Since the tunnel has a height of only 8 m, we must have that maximum height no more than 8 m: (-g/2)(u^2 sin^2(a)/g^2)+ u sin^2(a)/g= 8. Use that to find sin(a) as a function of u.

    The horizontal distance move in time t is d= ucos(a) t. By symmetry, the time until the particle hits the floor will be twice the time to the maximum height (so the time until the particle hits the ground again is, as you say 2u sin(a)/2. Use that, together with the value of a from the previous equation, to determine the range.
    Last edited by HallsofIvy; Jan 28th 2017 at 08:53 AM.
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  4. #4
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    Re: projectile

    Max range with an unrestricted height is normally achieved at a launch angle $\theta = 45^\circ$

    Max height, $h_{max} = \dfrac{[u\sin(45^\circ)]^2}{2g} = \dfrac{u^2}{4g}$

    To achieve max range with a launch angle of $45^\circ$, $h_{max} = 8 \implies \dfrac{u^2}{4g} \le 8 \implies u \le \sqrt{32g}$

    Note $R = \dfrac{u^2\sin(2\theta)}{g}$. With $u = \sqrt{32g}$ and launch angle $\theta = 45^\circ$, $R = 32 \, m$ which works in the problem's given formula,

    $R = 8 \cdot \sqrt{\dfrac{u^2 - 16g}{g}} = 8 \cdot \sqrt{\dfrac{32g - 16g}{g}} = 8 \cdot 4 = 32 \, m$


    If $u > \sqrt{32g}$, then the projectile would break the 8 m height restriction if launched at $45^\circ$, so the launch angle $\theta < 45^\circ$

    $\dfrac{(u\sin{\theta})^2}{2g} = 8 \implies \sin{\theta} = \dfrac{4\sqrt{g}}{u} \implies \cos{\theta} = \sqrt{1-\sin^2{\theta}} = \dfrac{\sqrt{u^2-16g}}{u}$

    $R = \dfrac{u^2\sin(2\theta)}{g} = \dfrac{u^2 \cdot 2\sin{\theta}\cos{\theta}}{g} = \dfrac{2u^2}{g} \cdot \dfrac{4\sqrt{g}}{u} \cdot \dfrac{\sqrt{u^2-16g}}{u} = \dfrac{\sqrt{u^2-16g}}{u} = \dfrac{8}{\sqrt{g}} \cdot \sqrt{u^2-16g} = 8 \sqrt{\dfrac{u^2-16g}{g}}$
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    Re: projectile

    Correction to the last line above ...

    $R = \dfrac{u^2\sin(2\theta)}{g} = \dfrac{u^2 \cdot 2\sin{\theta}\cos{\theta}}{g} = \dfrac{2u^2}{g} \cdot \dfrac{4\sqrt{g}}{u} \cdot \dfrac{\sqrt{u^2-16g}}{u} = \dfrac{8}{\sqrt{g}} \cdot \sqrt{u^2-16g} = 8 \sqrt{\dfrac{u^2-16g}{g}}$
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    Re: projectile

    i dont understand how you get cos theta =\cdot \dfrac{\sqrt{u^2-16g}}{u} i understand how you got sin theta though
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  7. #7
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    Re: projectile

    $\sin{\theta} = \dfrac{4\sqrt{g}}{u} \implies \sin^2{\theta} = \dfrac{16g}{u^2}$

    from the basic Pythagorean trig identity, $\cos^2{\theta} + \sin^2{\theta} = 1$, ...

    $\cos^2{\theta} = 1-\sin^2{\theta}$

    $\cos^2{\theta} = 1- \dfrac{16g}{u^2} = \dfrac{u^2}{u^2} - \dfrac{16g}{u^2} = \dfrac{u^2-16g}{u^2}$

    $\cos{\theta} = \sqrt{\dfrac{u^2-16g}{u^2}} = \dfrac{\sqrt{u^2-16g}}{\sqrt{u^2}} = \dfrac{\sqrt{u^2-16g}}{u}$
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