Question:

A projectile is given an initial velocity of (î+2ĵ) m/s, where î is along the ground and ĵ is along the vertical. If g=10 m/s², the equation of its trajectory is:

4y=2x-5x²

y=x-x²

y=2x-x²

4y=2x-x²

→u=î+2ĵ; ∴ucosθ=1; usinθ=2
x=ucosθt=t
y=usinθt-½gt²=2t-½gt²
Since g=10 m/s², y=2t-5t²
From x=t, t=x
Substituting in y=2t-5t², we get y=2x-5x²