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, $\left.g=10\, m / s ^{2}\right)$

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

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, $v_{\text {avg }}=\frac{O B}{t}=v_{0}$ or $O B=v_{0} t$
$\Rightarrow \,\sqrt{v_{0}^{2}+\left(v_{0}-10\right)^{2}}=v_{0} \sqrt{2}$
$ \Rightarrow \,v_{0}^{2}+\left(v_{0}-10\right)^{2}=2 v_{0}^{2}$
$\Rightarrow \, v_{0}-10=\pm v_{0} $
$ 2 v_{0}-10$
$ \Rightarrow v_{0}=5 ms ^{-1}$