A body is projected at an angle of 60 ° with the horizontal such that the vertical component of its initial velocity is 40   m   s - 1 . The magnitude of velocity of the projectile at one quarter of its time of flight is nearly (Acceleration due to gravity = 10   m   s - 2 )

Step-by-step solution

Suppose initial velocity is v and angle of projection θ is given 60 ° . Given, Vertical component of initial velocity v   sin   θ = 40   m   s - 1 ⇒     v = 46 . 19   m   s - 1 Time of flight T = 2   v   sin   θ g = 2 × 40 10 = 8   s Given, time t = T 4 = 2   s Suppose vertical component of velocity after time t is v 1 . Applying first eqaution of motion- v 1 = v   sin   60 ° - g t ⇒     v