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

" 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 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 (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 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 )

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) .

At a given instant of time the position vector of a particle moving in a circle with a velocity 3hat i - 4 hat j + 5 hat k is hat i + 9 hat j - 8 hat k . Its angular velocity at that time is:

A particle has initial velocity (3 hat i+ 4 hat j) and has acceleration (0.1 hat i+ 0.3 hat j) . Its speed after 10 s is.

A cannon fires its projectile with such an initial velocity and angle of projection that its range is R and the maximum height to which the projectile rises is H. Find the maximum range that can be obtained with the same initial velocity.

A cannon fires its projectile with such an initial velocity and angle of projection that its range is R and the maximum height to which the projectile rises is H.Find the maximum range that can be obtained with the same initial velocity.