Question

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

• A. $v_A=v_B=v_C$
• B. $v_A=v_B>v_C$
• C. $v_B>v_C>v_A$
• D. $v_A>v_B=v_C$

Answer 1

Answer: (A)

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 $A,v_A=\sqrt{u_A^2+2gh}$
For $B,v_B=\sqrt{u_B^2+2gh}$
For $C,v_C=\sqrt{u_C^2+2gh}$
As $u_A=u_B=u_C$ (Given)
$\therefore v_A=v_B=v_C$