Question

A ball is thrown with a kinetic energy E at an angle of $45{}^\circ$ with the horizontal in the earths gravitational field. The change in its potential energy at the highest point of its flight with respect to the starting point will be

• A. $+\frac{E}{\sqrt{2}}$
• B. $+\frac{E}{2}$
• C. $-\frac{E}{\sqrt{2}}$
• D. $-\frac{E}{2}$

Answer 1

Answer: (B)

$+\frac{E}{2}$

Answer 2

: At starting point, potential energy = 0 Maximum height attained by projectile $=\frac{{{u}^{2}}{{\sin }^{2}}\theta }{2g}$ $\therefore$ $h=\frac{{{u}^{2}}{{\sin }^{2}}45{}^\circ }{2g}$ or $h=\frac{{{u}^{2}}}{4g}$ $\therefore$ Potential energy gained $=mgh$ $PE=\frac{m{{u}^{2}}}{4}$ $PE=\frac{1}{2}\times \frac{1}{2}m{{u}^{2}}=\frac{1}{2}\times E$ or PE gained $=E/2$ .