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Thread: Projectile motion question

  1. #1
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    Post Projectile motion question

    I'm having trouble with a Yr.12 Maths C question on projectile motion (I hope this is the right section to post it)

    The question is: when a child standing on a horizontal path throws a ball, it leaves her hand from a point that is 90cm (0.9m) vertically above the path. The ball clears a 4.5-m-high wall that is 10.5m away from where it was thrown. SHow that "the least" velocity required for this to occur is the same as that acquired by a body falling 7.35m under gravity.

    I've tried this a few different ways- but keep getting stuck with multiple variables, etc.
    Can anyone help me out with this?

    my attempt, (note: X=theta)
    horizontal range= vcos(X)t and displacement(vertical)= vsin(X)t-4.9t^2 +0.9
    if x=10.5 and Y=3.6 at one point- we can sub these into the equations to get:

    Substitute and rearrange for v
    10.5= vcos(X)t {v=10.5/(cos(x)t)}
    4.5= vsin(x)t-4.9t^2+0.9 {v=(4.5+4.9t^2-0.9)/sin(x)t}

    therefore,
    10.5/(cos(x)t)=v=(4.5+4.9t^2-0.9)/sin(x)t
    simplifying,
    4.9t^2=10.5tan(X)-3.6

    this equation has 2 variables (the angle and time), is there any way to solve this {i recall my teacher mentioning it- but not in great detail}. Or is this the wrong way to attempt it?
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  2. #2
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    In a projectile motion, to reach a target, the velocity of projection will be minimum when the angle of projection is 45 degrees.
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  3. #3
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    Hello koopatroopa

    Welcome to Math Help Forum!
    Quote Originally Posted by koopatroopa View Post
    I'm having trouble with a Yr.12 Maths C question on projectile motion (I hope this is the right section to post it)

    The question is: when a child standing on a horizontal path throws a ball, it leaves her hand from a point that is 90cm (0.9m) vertically above the path. The ball clears a 4.5-m-high wall that is 10.5m away from where it was thrown. SHow that "the least" velocity required for this to occur is the same as that acquired by a body falling 7.35m under gravity.

    I've tried this a few different ways- but keep getting stuck with multiple variables, etc.
    Can anyone help me out with this?

    my attempt, (note: X=theta)
    horizontal range= vcos(X)t and displacement(vertical)= vsin(X)t-4.9t^2 +0.9
    if x=10.5 and Y=3.6 at one point- we can sub these into the equations to get:

    Substitute and rearrange for v
    10.5= vcos(X)t {v=10.5/(cos(x)t)}
    4.5= vsin(x)t-4.9t^2+0.9 {v=(4.5+4.9t^2-0.9)/sin(x)t}

    therefore,
    10.5/(cos(x)t)=v=(4.5+4.9t^2-0.9)/sin(x)t
    simplifying,
    4.9t^2=10.5tan(X)-3.6

    this equation has 2 variables (the angle and time), is there any way to solve this {i recall my teacher mentioning it- but not in great detail}. Or is this the wrong way to attempt it?
    You've started off with the correct equations, but you need to eliminate $\displaystyle t$ between them to get:
    $\displaystyle 4.5=10.5\tan\theta -\frac{4.9\cdot10.5^2\sec^2\theta}{v^2}+0.9$

    $\displaystyle \Rightarrow 3.6v^2=10.5v^2\tan\theta -4.9\cdot10.5^2(1+\tan^2\theta)$


    $\displaystyle \Rightarrow 4.9\cdot10.5^2\tan^2\theta -10.5v^2\tan\theta +3.6v^2+4.9\cdot10.5^2=0$

    Now here's the tricky bit. Thinking of this as a quadratic in $\displaystyle \tan\theta$, the minimum value of $\displaystyle v$ occurs when this equation has equal roots. This corresponds, of course, to there being just one value of $\displaystyle \theta$ for the ball to just clear the wall.

    So you now set the discriminant equal to zero to get:
    $\displaystyle (10.5v^2)^2 - 4\cdot4.9\cdot10.5^2(3.6v^2+4.9\cdot10.5^2)=0$

    $\displaystyle \Rightarrow(v^2)^2 - 4\cdot4.9\cdot3.6v^2-4\cdot4.9\cdot4.9\cdot10.5^2=0$, after dividing through by $\displaystyle 10.5^2$
    Using the quadratic formula on this equation, taking the positive root (since $\displaystyle v^2 > 0$) gives
    $\displaystyle v^2 = 144.06$
    Then, using $\displaystyle v^2 = 2gh$:
    When $\displaystyle g = 9.8$ and h = $\displaystyle 7.35$
    $\displaystyle v^2 = 144.06$
    and hence the required result.

    Grandad

    (Note: the angle of $\displaystyle 45^o$ only applies to the maximum range on a horizontal plane; and, of course, the top of the wall is not at the same height as the point of projection.)
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