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    Hi, could someone explain how to do part c please, I don’t understand it.
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    Use the formula you obtained in part a, but with d and h instead of x and y. Get an expression for d in terms of h and other stuff. You need to get rid of h so use the fact that h=1/2gT^2.
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    (Original post by Radioactivedecay)
    Use the formula you obtained in part a, but with d and h instead of x and y. Get an expression for d in terms of h and other stuff. You need to get rid of h so use the fact that h=1/2gT^2.
    But isn’t h=1/2gT^2 for when the projectile is launched horizontally like in part b? How can we use that in part c when it’s launced at an angle?
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    (Original post by nk2526)
    But isn’t h=1/2gT^2 for when the projectile is launched horizontally like in part b? How can we use that in part c when it’s launced at an angle?
    No. It works here because the intial speed is the same and so is the height above the ground, so the time taken has to be the same. Trust me, try it, it will work.
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    I get that it will work but I’m just trying to understand how. How will the time be the same if it’s launched at an angle though? Won’t the horizontal component then be ucosx which would be less than Just u, so it would take longer to travel the distance d
 
 
 
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