A particle A is projected from the ground with an initial velocity of 10 m s - 1 at an angle 60 ° with horizontal. From what height (in m ) directly above the point of projection of the particle A , should another particle B be projected horizontally with velocity 5 m s - 1 so that both the particles collide right before they strike the ground? [ g = 10 m s - 2 ]

Question

A particle A is projected from the ground with an initial velocity of 10 m s - 1 at an angle 60 ° with horizontal. From what height (in m ) directly above the point of projection of the particle A , should another particle B be projected horizontally with velocity 5 m s - 1 so that both the particles collide right before they strike the ground? [ g = 10 m s - 2 ],

MARKS App · Accepted Answer

Both the particles will meet at C , if their time of flight is the same. The time of flight of A is T = 2 u sin θ g = 2 × 10 × sin 6 0 ∘ 10 = 3 s For vertical downward motion of particle B from B to C We have h = 1 2 g T 2 = 1 2 × 10 × 3 2 = 15 m

Search any question & find its solution

Looking for more such questions to practice?