Three particles A, B, C are thrown from the top of a tower with the same speed. A is thrown straight up, B is thrown straight down and C is thrown horizontally. They hit the ground with speeds va,v_b and v_c and respectively.

Three particles A, B, C are thrown from the top of a tower with the sa

1.	Three particles A, B, C are thrown from the top of a tower with the same speed. A is thrown straight up, B is thrown straight down and C is thrown horizontally. They hit the ground with speeds va,v_b and v_c and respectively.
Answer» `v_a =v_b=v_c` `v_a` gt`v_b` gt`v_c` `v_a=v_b` gt `v_c` `v_a=v_b` lt`v_c` Solution :Here, since u of particle and H of the tower are equal for upward and DOWNWARD JOURNEY, final VERTICAL SPEED for A and B will be the same i.e `v_a =v_b` The particle C will have two velocities u as the horizontal velocity v as the verticle velocity. `:. v_c= sqrt(u^2 +v^2)` Hence `v_a` =`v_b` <`v_c`

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Three particles A, B, C are thrown from the top of a tower with the same speed. A is thrown straight up, B is thrown straight down and C is thrown horizontally. They hit the ground with speeds va,v_b and v_c and respectively.

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