Question

A particle is thrown up vertically with a speed $v_1$ in air. It takes time $t_1$ in upwards journey and $t_2(>t_1)$ in the downward journey and returns to the starting point with a speed $v_2$. Then,

• A. $v_1=v_2$
• B. $v_1<v_2$
• C. $v_1>v_2$
• D. data is insufficient

Answer 1

The correct option is C $v_1>v_2$

$v_1^2-2(g+a)h=0$
$\Rightarrow v_1=\sqrt{2(g+a)h}$
and $v_2^2=2(g-a)h$
$\Rightarrow v_2=\sqrt{2(g-a)h}$
$\therefore v_1>v_2$
There is loss of energy during the motion so it is clear that $v_1>v_2$