A body of mass 10 kg is projected at an angle of 45 ° with the horizontal. The trajectory of the body is observed to pass through a point 20 , 10 . If T is the time of flight, then its momentum vector, at time t = T 2 , is _____ . [Take g = 10 m s - 2 ]

Step-by-step solution

It is given that the projectile passes through 20 , 10 and angle of projection is θ = 45 ° . Using the trajectory equation, y = x tan θ - g x 2 2 u 2 cos 2 θ ⇒ 10 = 20 - 10 100 u 2 ⇒ u = 20   m   s - 1 Now the time of flight of projectile, T = 2 u sin θ g ⇒ T = 2 20 2 10 = 2 2   s ⇒ T 2 = 2   s Now velocity of the projectile at any time t can be written as, v → = u cos θ i ^ + u sin θ - g t j ^ ⇒ v → = 10 2 i ^ + ( 10 2