A projectile is given an initial velocity of `( hat (i) + 2 hat (j)) ms^(-1)` . The equation of its path is `(g= 10 ms^(-2))`

Options

A. `y=2x-5x^2`
B. `y= x-5x^2`
C. `4y=2x -5x^2`
D. `y=2x-25 x^2`

Answer

A

Explanation

Velocity of particle is `( hat i+ hat 2j) ms^(-1)` initially.
So, `u_x =1 ms^(-1)` and `u_y = 2ms^(-1)`
Also, `ax=0 and a_y=-10 ms^(-2)`
In time t,
Horizontal distance covered by projectile is
`x=u_x xx t=t`
And vertical distance covered by projectile is
`y=u_yt+1/2 a_y t^2`
`rArr y= 2t -5t^2`
Substiuting the value of t from Eq (i) in Eq (ii), we get
`y=2x-5x^2`