A boy is running along a straight horizontal road with a constant speed 5 ~ms ⁻¹ . While running he throws a stone with a velocity 30 ~ms ⁻¹ at an angle 60^ with the horizontal. Then the time of flight of the stone is: (Given g=10 ~ms ⁻² )

Step-by-step solution

The velocity of the stone is given with respect to the boy. Let the velocity of the boy be v _b = 5 i ms ⁻¹ . The velocity of the stone with respect to the boy is v _ s/b = 30 60^ i + 30 60^ j ms ⁻¹ . The velocity of the stone with respect to the ground is v _s = v _ s/b + v _b = (30 0.5 + 5) i + (30 3 2 ) j ms ⁻¹ = 20 i + 15 3 j ms ⁻¹ . The time of flight T for a projectile depends only on the vertical component of the initial velocity u_y and the acceleration due to gravity g . T = 2 u_y g = 2 15 3 10 = 30 3 10 =