The potential energy of a projectile at its highest point is3/4 of the value of its initial kinetic energy. Therefore its angle of projection is ..........

The potential energy of a projectile at its highest point is3/4 of the

1.	The potential energy of a projectile at its highest point is3/4 of the value of its initial kinetic energy. Therefore its angle of projection is ..........
Answer» `45^(@)` `30^(@)` `75^(@)` `60^(@)` Solution :`U= k_(0) - (3)/(4) k_(0)= (1)/(4) k_(0)` AtmaxheightU = K `K= K_(0) cos^(2) theta ` `(1)/(4) K_(0)= k_(0)cos^(2) theta` `:. , (1)/(4) =cos^(92)theta` `:. , (1)/(2) = cos theta` `:. , thetacos ^(-1)((1)/(2))` `theta= 60^(@)`