At a height `0.4 m` from the ground the velocity of a projectile in vector form is `vec v = (6 hat i+ 2 hat j) ms ^-1`. The angle of projection is

At a height of 0.4 m from the ground, the velocity of projectile in vector from is vec(u)=(6hat(i)+2hat(j)) (the x-axis is horizontal and y-axis is vertically upwards). The angle of projection is : (g = 10 m//s^(2))

The initial velocity of a projectile at the point of projection is 3hat i+4hat j ms^(-1) .Then the velocity of striking on the ground is

Find the angle of vector vec a = 6 hat i + 2 hat j - 3 hat k with x -axis.

Find the angle of vector vec a = 6 hat i + 2 hat j - 3 hat k with x -axis.

Find the angle of vector vec a = 6 hat i + 2 hat j - 3 hat k with x -axis.

Find the projection vector of vec b = hat i + 2 hat j + hat k along the vector vec a = 2 hat i + hat j + 2 hat k .

A body is projected from the ground with a velocity vec(v)= (3 hat(i) + 10 hat(j)) ms^(-1) . The maximum height attained and the range of the body respectively are (given g= 10 ms^(-2) )

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 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 )

Find the projection of the vector 7 hat i+ hat j-4 hat kin vec a=2 hat i+6 hat j+3 hat kdot