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Ask iit jee aieee pet cbse icse state board community Discussion Response Post to: Roorkee-trajectory
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[Post New]22 Nov 2007 18:42:44 IST
    Subject: Re:Roorkee-trajectory
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rhd92781 (686)

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Olaaa!! Perrrfect answer. 118  [166 rates]
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vx=u so dx/dt = u
integrating we get, x=ut+c
vy=wacoswt so dy/dt =wacoswt
integrating, y=asinwt + k
since at t=0, x=0 and y=0
so c=k=0
x=ut or t=x/u
y=asin(wx/u) is the trajectory
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