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    I'm really stuck on 3d, can anyone help? The answer to b is (24U^2)/g and the answer to c is 5U but I can't get the answer to d for some reason. (Sorry for attaching a picture twice but I did it from my phone and idk how to get rid it)
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    The horizontal velocity is fixed so you can work out what the vertical velocity had to be too make the magnitude of the velocity that.

    Then you know the starting vertical velocity and the vertical acceleration, so you can work out at what time the velocity equals this target value.

    Does that approach make sense? The key is splitting it into horizontal and vertical components of motion.
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    (Original post by selenacxoxo)
    I'm really stuck on 3d, can anyone help? The answer to b is (24U^2)/g and the answer to c is 5U but I can't get the answer to d for some reason. (Sorry for attaching a picture twice but I did it from my phone and idk how to get rid it)
    v_x=3U

    v_y=4U-gt

    therefore V^2=(v_x)^2+(v_y)^2

    \Rightarrow (U\sqrt{18})^2=(3U)^2+(4U-gt)^2

    Get a quadratic in t and U and complete the square on t
 
 
 
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