Question

A person standing near the edge of the top of a building throws two balls $A$ and $B$. The ball $A$ is thrown vertically upward and $B$ is thrown vertically downward with the same speed. The ball $A$ hits the ground with a speed $v_A$ and the ball $B$ hits the ground with a speed $v_B$. We have:

• A. $v_A>v_B$
• B. $v_A<v_B$
• C. $v_A=v_B$
• D. the relation between $v_a$ and $v_b$ depends on height of the building above the ground.

Answer 1

The correct option is C $v_A=v_B$

As no external force in acting, total energy of the system for Ball $A$

$\frac{1}{2}m{u}_A^2+mgh=\frac{1}{2}m{v}_A^2;v_A=\sqrt{u_A^2+2gh}$

For Ball $B$

$\frac{1}{2}m{u}_B^2+mgh=\frac{1}{2}m{v}_B^2;v_B=\sqrt{u_B^2+2gh}$

Since $u_A=u_B$

$v_a=v_B$.