A body projected with some velocity at an angle `45^@` with the horizontal from the origin in XY- plane passes through a point at (4,3) m . Its horizontal range is

Options

A. `10m `
B. `14 m`
C. `18 m`
D. `16 m`

Answer

D

Explanation

Equation of trajectory of projectile is
`y =x tan theta - (gx^2)/(2u^2 cos ^2 theta)`
Here, `theta = 45^2`
`rArr y=x-(gx^2)/(u^2)`
When y=3, x=4, so 3 `=4-(g(4^2))/(u^2)`
`rArr (16)/(u^2)=1`
`rArr u^2=16 g`
`or y=x-(x^2)/(16)` How at maximum x,y=0
Hence, `x-(x^2)/(16)=0 rArr x (1-x/16)=0`
`rArr x =0` (intial position) and x=16 m (final position)