A hall is projected from the ground with velocity ‘v’ such that its range is maximum

	COLUMN-I		COLUMN-II
I	Velocity at half of the maximum height	P	$\frac{\sqrt{3}v}{2}$
II	Velocity at the maximum height	Q	$\frac{v}{\sqrt{2}}$
III	Change in its velocity when it return to the ground	R	$v\sqrt{2}$
IV	Average velocity when it reaches the maximum height	S	$\frac{v}{2}\sqrt{\frac{5}{2}}$
		T	$2v$

 $I\to P;II\to Q;III\to R;IV\to S$
I $\to$  P, Range is maximum when the angle of projection is 450
 $H=\frac{V^2}{2g}{\sin }^2{45}^0=\frac{V2}{4g}$
Velocity, at half of the maximum height is ‘V’.
 $V_1^2=V^2{\sin }^2{45}^0-2g\frac{H}{2}$
 $V_1^2=\frac{V^2}{2}-\frac{V^2}{4}\Rightarrow V_1=\frac{\sqrt{3}v}{2}$
II $\to$ Q Velocity at the maximum height     $V_H=V\cos$   $\cos {45}^0$
 $\therefore V_H=\frac{V}{\sqrt{2}}$
III  $\to$  R Projection velocity at projection point, $\overline{V_2}=V\cos {45}^0\stackrel{\wedge }{i}+V\sin {45}^0\stackrel{\wedge }{j}$  at the point, when the body strikes the ground  $\overline{V_t}=V\cos {45}^0\stackrel{\wedge }{i}-V\cos {45}^0\stackrel{\wedge }{j}$
 $\Delta V=\overline{V_t}+(-{\overline{V}}_i)=2V\sin {45}^0-\stackrel{\wedge }{j}$
 $\left|\overline{\Delta V}\right|=2V\sin {45}^0=V\sqrt{2}$
IV  $\to$ S Average velocity =  $\frac{\text{total displacement}}{\text{total time}}$
Displacement  =  $\sqrt{{\left(\frac{R}{2}\right)}^2+H^2}$
 $$\begin{gathered} V_{ave}=\frac{\sqrt{\frac{R^2}{4}+H^2}}{\frac{V\sin \theta }{g}} \\ {V}_{ave}=\frac{\sqrt{R^2+4H^2}}{\frac{2V\sin \theta }{g}} \\ {V}_{ave}=\frac{\sqrt{{\left(\frac{V^2}{g}\right)}^2+4{\left(\frac{V^2}{4g}\right)}^2}}{\frac{\sqrt{2}V}{g}} \\ \therefore V_{ave}=\frac{V}{2}\sqrt{\frac{5}{2}} \end{gathered}$$