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A particle is projected from the bottom ...

A particle is projected from the bottom of an inclined plane of inclination `30^@`. At what angle `alpha ` (from the horizontal) should the particle be projected to get the maximum range on the inclined plane.

Text Solution

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To returns to the point of projection after one elastic collsion the particle must meet the plane at right angle to x-axis the plane.
`V_x-V_(x0)=(-gsinalpha)t_0`
`implies0-V_0costheta=(-gsinalpha)t_0`
`impliest_0=(V_0Costheta)/(gsinalpha)`…(i)
Motion of the particle alongs y aixs
`barV_(y)=barV_(y0)+(g cos alpha)t_0`
`impliesV_y=V_(y0)-(gcosalpha)t_0`
`implies-V_0sintheta=+V_0gcosalphat_0`
`implies` Equating (i)&(ii)
`implies(V_0costheta)/(gsinalpha)=(2V_0sintheta)/(g cosalpha)`
`impliescottehta=2tanalpha`.....(iii)
`impliescostheta=(2tanalpha)/(sqrt(1+4tan^(2)alpha))` & `sintheta=(1)/(sqrt(1+4tan^(2)alpha))`
Time of flight for to & fro motion of the perticle
`T=2t_0=(2V_0costheta)/(gsinalpha)`
`implies T=2t_0=(2V_0(2tanalpha)/(sqrt(1+4tan^(2)alpha)))/(gsinalpha)=(4V_0)/(gsqrt(1+3sin^(2)alpha))`
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