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For what angle of projection will a spark reach a maximum height of 2m?

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    • Thread Starter

    Can someone please explain how I do last part - (iii). I can do (i) and (ii).

    A firework is buried so that its top is at ground level and it projects sparks all at a speed of 8 ms^-1. Air resistance is neglected.

    Use g = 10 ms^{-1} in this question.

    See attached picture of firework. Basically sparks coming out from ground. But no spark with angle of exit less than 30 degrees.

    (i) Calculate the height reached by a spark projected vertically and explain why no spark can reach a height greater than this.

    U_v = 8 ms^{-1} (initial speed)

    S_v = U_vt + \frac{1}{2}at^2

    S_v = 8t -5t^2

    Max height reached when vertical velocity = 0 V_v = 0

    v_v = u_v + at

    0 = 8 - 10t

    t = \frac{4}{5} = 0.8

    S_v = 8(0.8) - 5(0.8^2) = 3.2 m

    3.2m is correct answer so that is fine.

    (ii) For a spark projected at 30 ^{\circ} to the horizontal over horizontal ground. Show that its height in metres t seconds after projection is  4t - 5t^2

    s_v = u_vt + \frac{1}{2}at^2

    u_v when speed is 8 and angle 30 degress is 8 x sin(30) = 4.

    s_v = 4t - 5t^2

     s_h = u_ht

    u_h = 8 \cos{30} = 6.93 ms^{-1}

    u_h = 6.93t

    Spark lands when s_v = 0 ie 0 = 4t - 5t^2

    t = \frac{4}{5} = 0.8, s_h = 6.93(0.8) = 5.5m

    But this next part I can't work out how to do.

    (iii) For what angle of projection will a spark reach a maximum height of 2m?

    s_v = 2 = u_v t - 5t^2.

    2 = 8 \sin{\theta} t - 5t^2.

    But there are two variables - t and the angle. So how do I work out angle???
    Attached Images

    You could always use v_v^2=u_v^2+2as_v, and say that at the maximum height v_v=0.

    Why would you want to find t? You aren't continuing from the same situation as (ii) - the angle is now the unknown you're trying to find, so you can't take it to be 30 (and if you did take it as 30 what other angle do you think you're trying to find?). The vertical component of u determines the maximum height, so find this value and equate it to 8sin(theta) to find the angle
    • Thread Starter

    (Original post by Arbolus)
    You could always use v_v^2=u_v^2+2as_v, and say that at the maximum height v_v=0.
    Ah thank you, got it.

    v_v^2 = u_v^2 + 2as_v

    Max height when v_v = 0

    0 = u_v^2 -40

    u_v = \sqrt{40}

    u_v = 8 sin{\theta}

    sin{\theta} = \frac{\sqrt{40}}{8}

    \theta = arcsin \frac{\sqrt{40}}{8} = 52
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