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A small object is thrown at an angle $45^{\circ}$ to the horizontal with an initial velocity $\mathbf{v}_0$. The velocity is averaged for first $\sqrt{2} \mathrm{~s}$ and the magnitude of average velocity comes out to be same as that of initial velocity, i.e. $\left|\mathbf{v}_{\mathrm{o}}\right|$. The magnitude $\left|\mathbf{v}_0\right|$ will be (take, $g=10 \mathrm{~m} / \mathrm{s}^2$ )
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Verified Answer
The correct answer is:
$5 \mathrm{~m} / \mathrm{s}$
Let object is at $B(x, y)$ after $t=\sqrt{2} \mathrm{~s}$
Then, $x=u_x \times t=v_0 \cos 45^{\circ} \times \sqrt{2}=v_0$
and $y=u_y t-\frac{1}{2} a_y t^2=v_0\left(\sin 45^{\circ}\right) \sqrt{2}-\frac{1}{2}(10) \times(\sqrt{2})^2$
$$
=\left(v_0-10\right) \mathrm{m}
$$
Displacement $O B$ of particle is
$$
O B=\sqrt{O A^2+A B^2}=\sqrt{v_0^2+\left(v_0-10\right)^2}
$$
So, $\quad v_{\text {avg }}=\frac{O B}{t}=v_0$ or $O B=v_0 t$
$$
\begin{array}{lrl}
\Rightarrow & \sqrt{v_0^2+\left(v_0-10\right)^2}=v_0 \sqrt{2} & \Rightarrow v_0^2+\left(v_0-10\right)^2=2 v_0^2 \\
\Rightarrow & v_0-10= \pm v_0 & 2 v_0=10 \Rightarrow v_0=5 \mathrm{~ms}^{-1}
\end{array}
$$
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