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a body is projected wid velocity u and at an angle x wid d horizontal.the body makes an angle of 30 degree at t = 2 sec and den after 1 sec it reaches maximum hieght.find u and x?
half time taken =usinx/g = 3sec therefore u sinx = 3g after 2 sec let its velocity be v , then v sin30 = u sinx -2g v/2 =3g-2g = g or v= 2g by putting the value of u sinx now as horizontal velocity remain constant, therefore u cosx = vcos30 u cosx = v√3/2 u cosx =√3 g by putting the value of v now dividing usinx by u cosx we get tanx =√3 or x = 60 degree then u = 3g / sinx = 20√3
half time taken =usinx/g = 3sec
therefore u sinx = 3g
after 2 sec let its velocity be v , then
v sin30 = u sinx -2g
v/2 =3g-2g = g or v= 2g by putting the value of u sinx
now as horizontal velocity remain constant, therefore u cosx = vcos30
u cosx = v√3/2
u cosx =√3 g by putting the value of v
now dividing usinx by u cosx we get tanx =√3 or x = 60 degree
then u = 3g / sinx = 20√3
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