Question

A small body is thrown at an angle to the horizontal with the initial velocity ${\overrightarrow{v}}_0$. Neglecting the air drag, find the mean velocity vector $⟨\overrightarrow{v}⟩$ averaged over the first $tsec$ and over the total time of motion.

• A. ${⟨\overrightarrow{v}⟩}_t={\overrightarrow{v}}_0-\frac{gt}{2}$, $⟨\overrightarrow{v}⟩={\overrightarrow{v}}_0-g\frac{\left({\overrightarrow{v}}_{0}g\right)}{g^2}$
• B. ${⟨\overrightarrow{v}⟩}_t={\overrightarrow{v}}_0-\frac{gt}{2}$, $⟨\overrightarrow{v}⟩={\overrightarrow{v}}_0+g\frac{\left({\overrightarrow{v}}_{0}g\right)}{g^2}$
• C. ${⟨\overrightarrow{v}⟩}_t={\overrightarrow{v}}_0$, $⟨\overrightarrow{v}⟩={\overrightarrow{v}}_0-g\frac{\left({\overrightarrow{v}}_{0}g\right)}{g^2}$
• D. None of these

Answer 1

The correct option is A ${⟨\overrightarrow{v}⟩}_t={\overrightarrow{v}}_0-\frac{gt}{2}$, $⟨\overrightarrow{v}⟩={\overrightarrow{v}}_0-g\frac{\left({\overrightarrow{v}}_{0}g\right)}{g^2}$

Initial velocity vector is $\overrightarrow{v_o}$.

Acceleration due to gravity is $-\overrightarrow{g}$ as it is in downward direction and has a constant value.

So displacement vector as function of time is, $\overrightarrow{s\left(t\right)}=\overrightarrow{v_{o}t}-\frac{1}{2}\overrightarrow{g}{t}^2$

So mean velocity vector after first $t$ sec is given as,

$<\overrightarrow{v}{>}_t=\overrightarrow{s\left(t\right)}/t=\overrightarrow{v_o}-\frac{1}{2}\overrightarrow{g}t$

Now, total time in projectile motion is given as,

$t_{total}=\frac{2(\overrightarrow{v_o}.\overrightarrow{g})}{g^2}$

Thus mean velocity vector, averaged over total time of motion is,

$<\overrightarrow{v}>=\overrightarrow{v_o}-\frac{1}{2}\overrightarrow{g}{t}_{total}=\overrightarrow{v_o}-\overrightarrow{g}\frac{(\overrightarrow{v_o}.\overrightarrow{g})}{g^2}$

Option A is correct.