devarshi-dt-logo

Question:

A projectile is given an initial velocity of ^i+2^j. The cartesian equation of its path is (Take g=10 m s⁻²).

y=2x-5x²

y=x-x²

y=2x-x²

y=2x-x²

Solution:

Given, →u=^i+2^j=ux^i+uy^j
Then, ux=1=ucosθ and uy=2=usinθ
∴tanθ=usinθ/ucosθ=2/1=2
The equation of trajectory of a projectile motion is
y=xtanθ−gx²/2u²cos²θ=xtanθ−gx²/2(ucosθ)²
∴y=x×2−10x²/2(1)²=2x−x².