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 horizontally. They hit the ground with speeds ${V_A},\,\,{V_B}$ and ${V_C}$ respectively.
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 horizontally. They hit the ground with speeds ${V_A},\,\,{V_B}$ and ${V_C}$ respectively.
${V_A} = {V_B} = {V_C}$
${V_A} = {V_B} > {V_C}$
${V_B} > {V_C} > {V_A}$
${V_A} > {V_B} = {V_C}$
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 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 $v^{2}-u^{2}=2 g h$
Therefore
For $A, v_{A}=\sqrt{u_{A}^{2}+2 g h}$
For $B, v_{B}=\sqrt{u_{B}^{2}+2 g h}$
For $C, v_{C}=\sqrt{u_{C}^{2}+2 g h}$
As $u_{A}=u_{B}=u_{C}$ (Given)
$\therefore \quad v_{A}=v_{B}=v_{C}$