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    Could someone provide me with a solution for this please, as it keeps cropping up in past papers and i have no idea how to do it.

    Thanks

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    I know one for sure is that u need to find theta angle.
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    yup its half of projectile motion diagram u will get. so use the formulas taught to u. projectile motions have two components vertical and horizontal and solve it for velocity. i remember it has formula for velocity
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    You do suvat in the vertical direction first. You know that s=2, u=0 and a=9.81, so you use s=ut+0.5at^2 to find the time taken to fall 2 metres.

    Then do a horizontal suvat, with s=5.1, a=0 (assuming no air resistance) and use the value of t you just calculated. Then use s=ut+0.5at^2 again to find out what u is.
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    I got D
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    As above, separate horizontal and vertical components: work out using suvat the time taken to drop 2 metres under the influence of gravity (his initial vertical velocity is zero). Then use this to work out his minimum initial horizontal speed speed: you know the distance needed, you just worked out the time and he can't accelerate in the air (I guess you can ignore air resistance).
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    work out time taken for motorcyclist to fall 2m using s=ut+1/2at^2 noting u=0
    divide the horizontal difference by this to find horizontal speed
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    answer is 8 m/sec
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    Thank you, the answer is D by the way.

    Samconly this is a help forum, you are meant to ask questions. Are you clueless or something?
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    The method below uses suvat, and the fact that vertical and horizontal components are independent. If you havent been taught this then you probably arent meant to know the method below.

    I would use s=ut+1/2at^2 for the vertical component and rearrange to t=(2s/a)^1/2 because u=0 it disappears. So t= (2x2/9.8)^1/2
    This will give you the time it takes for the guy to reach the ground.
    During this time he is moving to right with speed X. So t x X=5.1 so X=5.1/t. X=8m/s

    think thats right.
 
 
 

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