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Q.
Three particles $A, B$ and $C$ are thrown from the top of a tower with the same speed. $A$ is thrown up, $B$ is thrown down and $C$ is thrown horizontally. They hit the ground with speeds $v_A, v_B$ and $v_C$ respectively
When $A$ is thrown up with initial velocity $u_A$, it reaches the maximum height at zero velocity comes back to $P$ with the same initial $u_A. B$ has the initial velocity $u_B$. The vertical velocity for $C = 0. u_C$ is acting horizontally.
Using $v^2 - u^2 = 2gh$
Therefore
For $A, v_A = \sqrt{u^2_A + 2gh}$
For $B, v_B = \sqrt{u^2_B + 2gh}$
For $C, v_C = \sqrt{u^2_C + 2gh}$
As $u_A = u_B =u_C$ (Given)
$\therefore \:\:\:\:\: v_A = v_B =v_C$
