Question

A projectile is given an initial velocity of (i +2j) m/s, where is along the ground and j is along the vertical
If g = 10 m/s, the equation of its trajectory is
(JEE-Mains 2013. 4/120)
(1) y = x - 5x2
(2) y = 2x - 5x
(3) 4y = 2x - 5x (4) 4y = 2x - 25x2​

Answer 1

Answer:

it's option 1.

Initial Velocity (u) = i^+2j^ m/s

Hence horizontal component of initial velocity (ux) = i^ m/s

Therefore |ux| = 1m/s

Also angle of projection (0) = tan-1(uy/ux)= tan-1(2/1)

Therefore tan0 = 2

Trajectory or path of a projectile is given by: y = xtan0 - gx2 / 2ux2

Puting ux = 1 and tan0 = 2, we get,

y = 2x - gx2 / 2

or y = 2x - 5x2

Answer 2

Answer:

y=2x-5x2

Explanation:

we should also know formulas for this, so i am adding this to the above answer.

y=ax-bx2

a=tan tita

b=g/2u2cos2 tita