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.

Options

A. `v_a =v_b=v_c`
B. `v_a` gt`v_b` gt`v_c`
C. `v_a=v_b` gt `v_c`
D. `v_a=v_b` lt`v_c`

Answer

d

Explanation

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`