The direction of a projectile at a certain instant is inclined at an angle `prop` to the horizontal , after `t` second, it is inclined at an angle `beta`. Prove that the horizontal component of the velocity of the projectile is `("gt")/(tan prop - tan beta)`.

The direction of velocity of a projectile at a certain instant is inclined at angle 60° with the horizontal. After 1 second it is inclined at an angle 30°. If horizontal component of velocity of projection is g √3/k, then find k

The direction of velocity of a projectile at a certain instant is inclined at angle 60° with the horizontal. After 1 second it is inclined at an angle 30°. If horizontal component of velocity of projection is g √3/k, then find k

The direction of motion of a projectile at a certain instant is inclined at an angle 60^(@) to the horizontal. After sqrt(3) seconds it is inclined an angle 30^(@) . If the horizontal compound of velocity of projection is 3n (in m//sec) then value of n is : (take g = 10 m//s^(2))

The direction of motion of a projectile at a certain instant is inclined at an angle 60^(@) to the horizontal. After sqrt(3) seconds it is inclined an angle 30^(@) . If the horizontal compound of velocity of projection is 3n (in m//sec) then value of n is : (take g = 10 m//s^(2))

A projectile is thrown with velocity v at an angle theta with the horizontal. When the projectile is at a height equal to half of the maximum height,. The vertical component of the velocity of projectile is.