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

Step-by-step solution

Velocity of particle is ( i +2 j ) ms ⁻¹ initially. So, u_x=1 ~ms ⁻¹ and u_y=2 ~ms ⁻¹ Also, a_x=0 and a_y=-10 ~ms ⁻² In time t , Horizontal distance covered by projectile is And vertical distance covered by projectile is y=u_y t+ 1 2 a_y t^2 Substituting the value of t from Eq (i) in Eq (ii), we get y=2 x-5 x^2