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    Hey could someone help me with this?

    A stone is catapulted from ground level at an angle of 35degrees to the horizontal. If the stone hits the ground 25m from its point of projection. Find its speed of release and the time it is in the air.

    Help will be appreciated, thanks
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    Use the equations ?
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    this is your third projectiles question thread... stop trying to get people to do your homework for you.
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    i have two unknowns in all the equations
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    Think abt the vertical displacement.
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    When it hits the ground back, the vertical displacement is : __________________________
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    zero
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    (Original post by didgeridoo12uk)
    this is your third projectiles question thread... stop trying to get people to do your homework for you.
    well, am i not aloud to ask for help?, and also if u notice in only 1 of them someone gives me the actual answer, which i then reconfirmed. Sorry if im being rude, but im not
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    Exactly.
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    Think about using

    \large V = \large U + \small a \small t

    Connected with the vertical displacement, and the fact that at the maximum height  \large V = 0

    The time for this will be half the overall time
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    (Original post by hassantahir)
    well, am i not aloud to ask for help?, and also if u notice in only 1 of them someone gives me the actual answer, which i then reconfirmed. Sorry if im being rude, but im not
    you can ask for help, but show all your working so far so we can see where you're going wrong
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    (Original post by CookieGhoul)
    Think about using

    \large V = \large U + \small a \small t

    Connected with the vertical displacement, and the fact that at the maximum height  \large V = 0

    The time for this will be half the overall time
    I still have two unknowns, no? U and T
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    So far i have got : t= 25/(Ucos35) i dont know if that is right though
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    (Original post by hassantahir)
    So far i have got : t= 25/(Ucos35) i dont know if that is right though
    Yes now work out t using the vertical displacement and use simultaneous equations to work out the initial velocity and in turn t.
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    (Original post by hassantahir)
    So far i have got : t= 25/(Ucos35) i dont know if that is right though

    Yea thats for the horizontal displacement. Now work the vertical displacement
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    How about this:

     \large V = \large U + \small a \small t



     0 = \large U \sin 35 + (-10)(\frac {\small t}{2})


    How about now ?
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    t=Usin35/4.9
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    Right. Can you solve it now ?
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    (Original post by hassantahir)
    t=Usin35/4.9
    yep that's what I got, now substitute for t and solve for U
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    25/Ucos35=Usin35/4.9
 
 
 
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