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Math Help - calculation distance

  1. #1
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    calculation distance

    A batsman strikes a cricket ball at a height of 1.5m above the ground, giving it an initial
    speed of 29 ms^{-1}
    at an angle of 30o measured upwards from the horizontal.



    (a) How far does the ball travel horizontally before touching the ground?
    (b) A fielder moves to try to catch the ball. If he can catch a ball that is at a height
    2.75m or below, where should he position himself?

    for part a)

    I need to work out the distance. but not sure what formula to use, and the fact that's its inclined at an angle. ?

    b) not sure either.



    any help appreciated!



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  2. #2
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    Re: calculation distance

    Quote Originally Posted by Tweety View Post
    A batsman strikes a cricket ball at a height of 1.5m above the ground, giving it an initial
    speed of 29 ms^{-1}
    at an angle of 30o measured upwards from the horizontal.



    (a) How far does the ball travel horizontally before touching the ground?
    (b) A fielder moves to try to catch the ball. If he can catch a ball that is at a height
    2.75m or below, where should he position himself?

    for part a)

    I need to work out the distance. but not sure what formula to use, and the fact that's its inclined at an angle. ?

    b) not sure either.



    any help appreciated!



    for (a) find the initial x and y velocities from the geometry then solve the equation of motion in the vertical direction for y=0. That gives you the time at which the ball hits the ground. Then use that time and the horizontal velocity to find the horizontal distance travelled.

    (b) is essentially the same problem but solve now for y=2.75. There will be two solutions, one on the way up and one on the way down. You want the one on the way down. Again use that t and the horizontal velocity to calculate where the fielder should be.
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  3. #3
    MHF Contributor

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    Re: calculation distance

    Separate into horizontal and vertical components. If the original horizontal velocity is u and there is no force to change the velocity, in time t, the object will have moved distance ut horizontally. If the original vertical speed is v and there is acceleration -g, then in time t it will have moved -(1/2)gt^2+ vt vertically. Here, the initial horizontal speed is 29 cos(30) and the initial vertical speed is 29 sin(30).
    Last edited by HallsofIvy; February 25th 2014 at 06:54 AM.
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