Question

Two bodies are projected vertically upwards from one point with the same initial velocities $v_{0}m/s$. The second body is thrown $\tau s$ after the first. The two bodies meet after time

• A. $\frac{v_0}{g}-\frac{\tau }{2}$
• B. $\frac{v_0}{g}+\tau$
• C. $\frac{v_0}{g}+\frac{\tau }{2}$
• D. $\frac{v_0}{2g}-\tau$

Answer 1

The correct option is C $\frac{v_0}{g}+\frac{\tau }{2}$

Height of first body after time $t,h_1=v_{0}t-\frac{1}{2}g{t}^2$
Height of the second body after time $(t-\tau )$,
$h_2=v_0(t-\tau )-\frac{1}{2}g(t-\tau {)}^2$
If they meet after time $t$, $h_1=h_2$
$\Rightarrow t=\frac{v_0}{g}+\frac{\tau }{2}$
The correct option is (c).