Question

A ball whose kinetic energy is $E$, is projected at an angle of $45^{\circ}$ to the horizontal. The kinetic energy of the ball at the highest point of its flight will be :

• A. $E$
• B. $E / \sqrt{2}$
• C. $E / 2$
• D. zero

Answer 1

Answer: (C)

At the highest point of its flight, vertical component of velocity is zero and only horizontal component is left which is $u_{x}=u \cos \theta$
Given $\theta=45^{\circ}$
$\therefore u_{x}=u \cos 45^{\circ}=\frac{u}{\sqrt{2}}$
Hence, at the highest point kinetic energy
$E^{\prime}=\frac{1}{2} m u_{x}^{2} =\frac{1}{2} m\left(\frac{u}{\sqrt{2}}\right)^{2}$
$=\frac{1}{2} m\left(\frac{u^{2}}{2}\right)$
$=\frac{E}{2} \left(\because \frac{1}{2} m u^{2}=E\right)$