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    Can someone help with this question? Thanks
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    Without trying it myself, I'd tackle it like this (I don't know the dimensions ofthe ball so am assuming it can travel 3m up before hitting the roof).

    (1) What vertical speed wiill give you a maximum height of 3m?

    (2) how long will it take for a ball launched at this vertical speed to reach the ground again?

    (3) Given the total velocity in the Q and the vertical component you've jsut calculated, find the horizontal component

    (4) Now, using the time you calculated in 82) and the horizontal speed in (3) you should be able to find how far it goes.


    I'm making the assumption here that at any smaller angle of launch will give a smaller range and any greater angle of launch will mean the initial vertical speed is such that it hits the roof.
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    it is worth remembering that maximum range is always achieved by a 45° launch angle.
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    (Original post by the bear)
    it is worth remembering that maximum range is always achieved by a 45° launch angle.
    It's also worth noting that in this case a 45º angle will mean it hits the roof.
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    (Original post by phys981)
    It's also worth noting that in this case a 45º angle will mean it hits the roof.
    this is true :teehee:
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    Thank you guys!!
 
 
 
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