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Math Help - projectiles?

  1. #1
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    projectiles?

    A golfer hits a ball from point A with initial velocity 24.5 m/s at and angle \alpha above the horizontal, where sin \alpha = 0.8. The ball just clears the tops of the two trees. The tops of the trees are both 14.7m above the level of a and are horizontal distance dm apart.

    a i) Find the time taken for the ball to reach the top of the first tree.

    ii) Determine the value of d.


    Ok i have worked this question out but i'm not confident the answer is right as i found it very hard. This is what i did.

    v = 24.5
    sin \alpha = 0.8
    tree h = 14.7

    so using trig i get 14.7/hypotenuse = 0.8

    hypotenuse = 14.7/0.8 = 18.475

    cos \alpha = adjacent/18.375

    18.375^2 = 14.7^2 + a^2

    a= \sqrt{18.375^2 - 14.7^2}

    \therefore a = 11.025

    cos\alpha = \frac{11.025}{18.375}

    cos \alpha = 0.6


    x = (vsin\alpha)t - 4.9t^2

    (24.5*0.8)t - 4.9t^2 = 14.7

    19.6t - 4.9t^2 = 14.7

    t^2 - 4t + 3 =0

    (t-3)(t-1)

    t = 3 or t=1

    so t=1 at top of first tree.


    time to second tree = 3

    (vcos\alpha)t = x_2 where x_2 = distance to second tree

    24.5 * 0.6 * 3 = x_2

    x_2 = 44.1

    x_2 - x_1 = d

    44.1 - 11.025 = 33.075

    could anybody confirm this is done right? or tell me where i have gone wrong if not?
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  2. #2
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    Quote Originally Posted by djmccabie View Post
    A golfer hits a ball from point A with initial velocity 24.5 m/s at and angle \alpha above the horizontal, where sin \alpha = 0.8. The ball just clears the tops of the two trees. The tops of the trees are both 14.7m above the level of a and are horizontal distance dm apart.

    a i) Find the time taken for the ball to reach the top of the first tree.

    ii) Determine the value of d.


    Ok i have worked this question out but i'm not confident the answer is right as i found it very hard. This is what i did.

    v = 24.5
    sin \alpha = 0.8
    tree h = 14.7

    so using trig i get 14.7/hypotenuse = 0.8

    hypotenuse = 14.7/0.8 = 18.475

    cos \alpha = adjacent/18.375

    18.375^2 = 14.7^2 + a^2

    a= \sqrt{18.375^2 - 14.7^2}

    \therefore a = 11.025

    cos\alpha = \frac{11.025}{18.375}

    cos \alpha = 0.6
    Uhh, sin^2(\alpha)+ cos^2(\alpha)= 1
    .8^2+ cos^(\alpha)= .64+ cos^2(\alpha)= 1
    cos^(\alpha)= 1- .64= .36
    cos(\alpha)= \pm .6
    and since \alpha must be in the first quadrant, cos(\alpha)= .6.


    x = (vsin\alpha)t - 4.9t^2
    You are using x to represent height? Peculiar, but okay.

    [tex] (24.5*0.8)t - 4.9t^2 = 14.7

    19.6t - 4.9t^2 = 14.7

    t^2 - 4t + 3 =0

    (t-3)(t-1)

    t = 3 or t=1

    so t=1 at top of first tree.


    time to second tree = 3

    (vcos\alpha)t = x_2 where x_2 = distance to second tree

    24.5 * 0.6 * 3 = x_2[/quote]
    Okay, the horizontal speed is constant so this is correct

    x_2 = 44.1

    x_2 - x_1 = d

    44.1 - 11.025 = 33.075

    could anybody confirm this is done right? or tell me where i have gone wrong if not?
    It looks good to me.
    Last edited by mr fantastic; May 16th 2009 at 08:11 PM. Reason: Fixed first line of latex
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  3. #3
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    i got it right..? feel like i've finally made some progress!
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