A ball having kinetic energy (KE), is projected at an angle of \(60^\circ\) from the horizontal. What will be the kineti…

\({K}_{\cdot E}\cdot =\frac{1}{2}m{u}^{2}\\ \text{ At maximum height }\\ K\cdot {E}_{1}=\frac{1}{2}m(u\cos \theta {)}^{2}\\ K\cdot {\epsilon }_{1}=\frac{1}{2}m{u}^{2}{\cos }^{2}60^\circ \\ K\cdot {\epsilon }_{1}=K\cdot E\cdot x\frac{1}{4}\\ K\cdot {\epsilon }_{1}=\frac{K\cdot E\cdot }{4}\)