Question

A small object is thrown at an angle $45^{\circ }$ to the horizontal with an initial velocity $v_0$ The velocity is averaged for first $\sqrt{2}s$ and the magnitude of average vclocity comcs out to bc same as that of initial velocity, i.e. $\left|v_0\right|$. The magnitude $\left|v_v\right|$ will be (take, $g=10m/s^2)$

• A. $3m/s$
• B. $3\sqrt{2}m/s$
• C. $4m/s$
• D. $5m/s$

Answer 1

Answer: (D)

Let object is at $B(x,y)$ after $t=\sqrt{2}s$
Then, $x=u_x\times t=v_0\cos 45^{\circ }\times \sqrt{2}=v_0$
and $y=u_{y}t-\frac{1}{2}{a}_y{t}^2=v_0\left(\sin 45^{\circ }\right)\sqrt{2}-\frac{1}{2}(10)\times (\sqrt{2}{)}^2$
$=\left(v_0-10\right)m$
Displacement $OB$ of particle is
$OB=\sqrt{O{A}^2+A{B}^2}=\sqrt{v_0^2+{\left(v_0-10\right)}^2}$
So, $v_{\text{avg }}=\frac{OB}{t}=v_0$ or $OB=v_{0}t$
$\Rightarrow \sqrt{v_0^2+{\left(v_0-10\right)}^2}=v_0\sqrt{2}$
$\Rightarrow v_0^2+{\left(v_0-10\right)}^2=2v_0^2$
$\Rightarrow v_0-10=\pm v_0$
$2v_0-10$
$\Rightarrow v_0=5m{s}^{-1}$