An object is thrown along a direction inclined at an angle of `45^(@)` with the horizontal direction. The horizontal range of the particle is equal to

Options

A. vertical height
B. twice the vertical height
C. thrice the vertice height
D. four times the vertical height

Answer

D

Explanation

`(R)/(H) = ((u^(2)sin 2theta)/(g))/((u^(2)sin^(2)theta)/(2g)) = (2sin theta cos theta)/(g) xx (2g)/(sin^(2)theta) = 4 cot theta`
`rArr R = 4H cot theta` if `theta = 45^(@)` then `R = 4H cot (45^(@)) = 4H`