A particle of mass m is projected at an angle with horizontal with kinetic energy E. the potential energy at the top of its trajectory is `E/2`. Find the range.

Options

A. `E/(mg)`
B. `E/(2mg)`
C. `(2E)/(mg)`
D. `E/(mgsqrt(2))`

Answer

3

Explanation

KE lost `=E/2 =1/2 (1/2 mV^(2))=1/2m(V/(sqrt(2)))^(2)`
Thus velocity at top becomes `V/(sqrt(2))`. If V is the projection velocity
Thus angle at which particle is projected
Range `=(u^(2) sin 2 theta)/g=(V^(2))/g sin90^(@)`
`=1/g((2E)/m) =(2E)/(mg)`