From a building two balls $\mathrm{A}$ and $\mathrm{B}$ are thrown such that $\mathrm{A}$ is thrown upwards…
From a building two balls $\mathrm{A}$ and $\mathrm{B}$ are thrown such that $\mathrm{A}$ is thrown upwards $\mathrm{A}$ and $\mathrm{B}$ downwards (both vertically). If $\mathrm{v}_{\mathrm{A}}$ and $\mathrm{v}_{\mathrm{B}}$ are their respective velocities on reaching the ground, then
$\mathrm{V}_{\mathrm{B}}>\mathrm{V}_{\mathrm{A}}$
$\mathrm{V}_{\mathrm{A}}=\mathrm{V}_{\mathrm{B}}$
$\mathrm{V}_{\mathrm{A}}>\mathrm{V}_{\mathrm{B}}$
their velocities depend on their masses
Solution
As the ball moves down from height ' $h$ ' to ground the P.E at height ' $h$ ' is converted to K.E. at the ground (Applying Law of conservation of Energy)
Hence, $\frac{1}{2} m_A v_A^2=m_A g h_A$ or $v_A=\sqrt{2 g h_A}$; Similarly, $v_B=\sqrt{2 g h}$ or $v_A=v_B$