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Give the expression for the range of the...

Give the expression for the range of the particle. For what value of angle of projection of the particle is the range maximum ?

Text Solution

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(a) Range of the projectile, `R=(v_0^2sin2theta_0)/g=2/g(v_0)_x(v_0)_y`
(b) If `theta_0=45^(@),sin2(45)^(@)=1 " ":." "R_("max")=(v_0^2)/(g)`. Hence for `theta_0=45^(@)`, the range is maximum
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