" A projectile is given an initial velocity "hat i+2hat j" .The cartesian equation of its path is "[" take "g=10m/s^(2)]

A projectile is given an initial velocity hati+2 hatj .The cartesin equation of its path is (g=10ms^(-2))

A projectile is given an initial velocity of (hat(i)+2hat(j)) The Cartesian equation of its path is (g = 10 ms^(-1)) ( Here , hati is the unit vector along horizontal and hatj is unit vector vertically upwards)

A projectile is given an initial velocity 3hat i+4hat jm/sec Its maximum height is

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 :

A projectile is given an initial velocity of (hati+2hatj)m/s , where hati is along the ground and hatj is along the vertical. If g=10m/s^-2 , the equation of its trajectory is

A body is thrown with an initial velocity "sqrt(3)hat i+1hat j" .Its angle of projection is

A projectile is thrown with an initial velocity of v = a hat i + b hat j . If the range of projectile is double the maximum height reached by it. Find the ratio (b)/(a) .

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

A projectile is projected in air with initial velocity vec v=(3 hat i + 4 hat j) m/s from the origin. The equation of trajectory of the projectile is given as (g=-10 hat j m/s^2) (neglect air resistance )