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    A cricketer hits the ball on the half volley, that is when the ball is at ground level. The ball leaves the ground at an angle of 30 degrees to the horizontal and travels towards a fielder standing on the boundary 60m away.
    i) Find the initial speed of the the ball if it hits the ground for the first time at the fielders feet.
    ii) Find the initial speed of the ball if it is at a height of 3.2m when it passes over the fielder's head.

    For part i) I know the horizontal range is given which is S=ut in this case 60m, I also know sin30 = 1/2 and cos30 = √3/2 but that's all I don't know how to approach this question. Any help would be greatly appreciated, thanks.
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    (Original post by Revision0161)
    A cricketer hits the ball on the half volley, that is when the ball is at ground level. The ball leaves the ground at an angle of 30 degrees to the horizontal and travels towards a fielder standing on the boundary 60m away.
    i) Find the initial speed of the the ball if it hits the ground for the first time at the fielders feet.
    ii) Find the initial speed of the ball if it is at a height of 3.2m when it passes over the fielder's head.

    For part i) I know the horizontal range is given which is S=ut in this case 60m, I also know sin30 = 1/2 and cos30 = √3/2 but that's all I don't know how to approach this question. Any help would be greatly appreciated, thanks.
    Hi. Try the following steps...

    1. If you let the initial speed of the ball be u, then work out expressions for the horizontal and vertical components of this.

    2. Resolving horizontally, work out an expression for the time taken to travel 60 metres

    3. Resolving vertically, work out an expression for the distance travelled using s=ut + (1/2)at2
    .
    - For part A, this distance will be zero as the ball hits the ground. You can substitute for t using your expression from part 2 and solve this for u

    - For part B, this distance will be 3.2m. You can substitute the same expression for t and solve for u.
    Spoiler:
    Show
    1. Resolving horizontally: 60m = ucos(30) x t; t = 60/(u x cos(30))

    2. Resolving vertically: s = usin(30) x t - 0.5gt2

    For part A: solve 0 = usin(30) x t - 0.5gt2 for u, substituting for t
    For part B: solve 3.2 = usin(30) x t - 0.5gt2 for u, substituting for t
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    Hi, thanks for your response.
    I followed your instructions and managed to solve it easily, however I have one question will this method involving expressions always work when I am trying to find the initial speed? Thanks again
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    (Original post by Revision0161)
    Hi, thanks for your response.
    I followed your instructions and managed to solve it easily, however I have one question will this method involving expressions always work when I am trying to find the initial speed? Thanks again
    Hi. Sorry about the delay.

    Yes, the method will always work. It relies on the fact that the horizontal speed is constant, so you can find an expression for t, which is then substituted into the other equation.
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    Thanks again for your help, by the way do you know any websites or resources that I can access to get additional help for M1. I have already tried exam solutions.
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    (Original post by Revision0161)
    Thanks again for your help, by the way do you know any websites or resources that I can access to get additional help for M1. I have already tried exam solutions.
    Sorry, I don't. It may be worth searching on Youtube to see if there is anything there.

    The links on the right may lead somewhere, e.g. I found this one via another page.

    M1 Mechanics - Pulleys - Connected Particles (Full 1 hour tutorial AQA, Edexcel)
    https://www.youtube.com/watch?v=r73j4prNAug
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    Okay, thanks a lot!
 
 
 
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