First slide

Conservation of mechanical energy

Question

A body has a kinetic energy E when projected at an angle of projection for maximum range. Its kinetic energy at its highest point of its path will be

Moderate

A

E
B

$\frac{E}{2}$
C

$\frac{E}{\sqrt{2}}$
D

zero

Solution

Maximum range is obtained at $45^{o}$
$K i n e t i c e n e r g y = E = \frac{1}{2} m u^{2}, - - - - (1) h e r e m i s m a s s o f b o d y, u i s v e l o c i t y o f t h e b o d y$
At highest point of projectile motion velocity v=u cos $θ$ , v=u cos $45^{o}$ = $\frac{u}{\sqrt{2}}$
$\begin{array}{l} A t t h e h i g h e s t p o i n t K E = K E^{'} = \frac{1}{2} m v^{2} = \frac{1}{2} m {(\frac{u}{\sqrt{2}})}^{2} \\ K E^{'} = \frac{(\frac{1}{2} m u^{2})}{2} = \frac{E}{2}, o b t a i n e d o n s u b s t i t u t i n g e q u a t i o n (1) \end{array}$

Get Instant Solutions

When in doubt download our app. Now available Google Play Store- Doubts App

Recieve an sms with download link

Doubts App

OR

SCAN ME

Doubts App

Doubts App