Question

A particle is projected at $60^{°}$ to the horizontal with an energy $E$. The kinetic energy and potential energy at thehighest point are

• A. $\left(\frac{E}{2},\frac{E}{2}\right)$
• B. $\left(\frac{3E}{4},\frac{E}{4}\right)$
• C. $(E,0)$
• D. $\left(\frac{E}{4},\frac{3E}{4}\right)$

Answer 1

Answer: (D)

Initial KE, $E=\frac{1}{2}m{u}^2...\left(i\right)$
At the highest point, velocity $\upsilon =ucos60^{°}=\frac{u}{2}$
KE at highest point $=\frac{1}{2}m{\upsilon }^2=\frac{1}{2}m{\left(\frac{u}{2}\right)}^2$
$=\frac{1}{4}\left(\frac{1}{2}m{u}^2\right)=\frac{E}{4}$ (Using (i))
PE at highest point $=mg\times \frac{u^{2}si{n}^{2}60^{°}}{2g}$
$=\frac{1}{2}m{u}^2{\left(\frac{\sqrt{3}}{2}\right)}^2=\frac{3}{4}E$