A projectile is given an initial velocity of ` ( hat(i) + 2 hat (j) ) m//s`, where ` hat(i)` is along the ground and `hat (j)` is along the vertical . If ` g = 10 m//s^(2) `, the equation of its trajectory is :

Options

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

Answer

B

Explanation

` vec(u) = hat(i) + 2hat(j) = u_(x)hat(i)+ u_(y)hat(j) rArr u cos theta = 1 , u sin theta = 2 `
` y = x tan theta - 1/2 (gx^(2))/(u_(x)^(2))`
`:. y = 2x - 1/2 gx^(2) = 2x - 5x^(2) `