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Thread: A projectile fired at a speed of u and an angle of x to the horizontal

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    A projectile fired at a speed of u and an angle of x to the horizontal

    A projectile fired at a speed of u and an angle of x to the horizontal, reaches a
    maximum height of H. When the same projectile is fired again at the same
    angle with a new higher initial speed it reaches a new maximum height of 2H.

    What is the new initial velocity in terms of u.
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    Re: A projectile fired at a speed of u and an angle of x to the horizontal

    The new initial velocity should be 'root 2 times u'. Tell me if there's something wrong with it or you don't understand a part of it: https://imgur.com/PuPIDE5
    Last edited by tazhossain99; 6 Dec 2017 at 7:32 PM.
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    Re: A projectile fired at a speed of u and an angle of x to the horizontal

    That's for the new uy. But for the new initial velocity, which is root of uy2 + ux2, I got u times root (sin squared x plus 1).
    Not sure if it's correct, so a confirmation would be appreciated.

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    Re: A projectile fired at a speed of u and an angle of x to the horizontal

    Quote Originally Posted by Powereaper View Post
    That's for the new uy. But for the new initial velocity, which is root of uy2 + ux2, I got u times root (sin squared x plus 1).
    Not sure if it's correct, so a confirmation would be appreciated.






    Last edited by InteGrand; 9 Jan 2018 at 10:58 AM.

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    Re: A projectile fired at a speed of u and an angle of x to the horizontal

    Quote Originally Posted by Powereaper View Post
    That's for the new uy. But for the new initial velocity, which is root of uy2 + ux2, I got u times root (sin squared x plus 1).
    Not sure if it's correct, so a confirmation would be appreciated.

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    Re: A projectile fired at a speed of u and an angle of x to the horizontal

    Oh,got it now. I thought horizontal velocity was the same, until I realised the question didn't mention that. Thanks for the help.

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