A body is projected at an angle of `60^(@)` with ground. It covers a horizontal distance of 100m. If the same body is projected at `60^(@)` with vertical with same velocity, the new range is

Options

A. `50m`
B. `100m`
C. `200m`
D. `150m`

Answer

B

Explanation

Range, `R=(u^(2)sin 2theta)/g`
For same v, `R prop 2 theta`
From `theta=600`with horizontal, R is given as 100m when
`R propsin2xx60^(@) prop sin 120^(@)......(i)`
For `theta=600` with vertical
`R' prop sin 2(90^(@)-60^(@))`
i.e., `R' prop sin 60^(@)`
Using (i) and (ii) `(R')/R=(sin 60^(@))/(sin 120^(@))=1/1`
i.e, `R'=R=100m`