A projectile is given an initial velocity of `(hati+2hatj)ms^(-1)` where, is along the ground and `hatj` is along the vertical. If `g=10ms^(-1)` what will be the equation of its trajectory ?

Options

A. ` y = 2x - 5 x ^ 2 `
B. ` y = x - 5x ^2 `
C. ` 4y = 2x - 5x ^ 2 `
D. ` y = 2x + 5x ^ 2 `

Answer

D

Explanation

Given initial velocity of the particle, `u=ghati+2hatj`
`therefore u_(x)=1m//s and u_(x)=2m//s`
Also `x=u_(x)t therefore x=t .......(1)`
and `y=u_(y)t-1/2gt^(2)`
`and y=2t-1/2 xx 10 xx t^(2)=2t-5t^(2) .....(2)`
From (1) and (2) Equation of trajectory is `y=2x-5x^(2)`