Question

A ball is projected from the ground with velocity $v$ such that its range is maximum. Match Column I entries with the correct entries in Column II.
$\begin{array}{cc}\text{Column I}& \text{Column II}\\ \text{i.Velocity at half of the maximum height}& a.\sqrt{2}v\\ \text{ii.Velocity at the maximum height}& b.\sqrt{3}v/2\\ \text{iii. Change in velocity when its returns to the ground}& c.\frac{v}{\sqrt{2}}\\ \text{iv. Average velocity when its reaches the maximum height}& d.\frac{v}{2}\sqrt{\frac{5}{2}}\end{array}$

• A. i – b ii – c iii – a iv – d
• B. i – a ii – c iii – d iv – b
• C. i – b ii – a iii – c iv – d
• D. i – a ii – c iii –b iv – d

Answer 1

The correct option is A i – b ii – c iii – a iv – d

For an projectile motion, range is maximum when the body is projected at an angle of ${45}^{\circ }$.
Thus $\theta ={45}^{\circ }$
Maximum height, $\begin{array}{}h=\frac{u^{2}si{n}^2\theta }{2g}=\frac{v^2(1{)}^2}{2\times 2g}=\frac{v^2}{4g}& \text{(1)}\end{array}$
At half of maximum height$\left(h^{\prime }\right)$, let the vertical component of velocity be $V_y^{\prime }$, then
$\begin{array}{}{V}_y^{\prime 2}-u_y^2=2g{h}^{\prime }& \text{(2)}\end{array}$
$u_y=vsin{45}^{\circ }=\frac{v}{\sqrt{2}}$
Thus equation $\left(2\right)$ will be,
$V_y^{\prime 2}-\frac{v^2}{2}=2g\frac{h}{2}$
Substituting the value $h$ from $\left(1\right)$,
$V_y^{\prime 2}-\frac{v^2}{2}=g\frac{v^2}{4g}$
$\Rightarrow V_y^{\prime }=\frac{v}{2}$
Since there is no acceleration in horizontal direction, therefore horizontal component of velocity remains component,
$V_x=vcos{45}^{\circ }=\frac{v}{\sqrt{2}}$
hence net velocity at half of maximum height, $V=\sqrt{V_x^2+V_y^2}=\sqrt{\frac{v^2}{2}+\frac{v^2}{4}}$
$\Rightarrow V=\frac{\sqrt{3}v}{2}$
Velocity at maximum height will have only horizontal component which remains constant due to absence of acceleration in horizontal direction. Hence velocity at maximum height is $\frac{v}{\sqrt{2}}$
Components of velocity of ball at starting and end position will have the direction as shown in the figure below.

Its evident from figure that change in velocity occurs due to change in vertical component only,
Change in velocity $=-V_y-V_y=-2V_y=-2vcos{45}^{\circ }=\sqrt{2}v$
Average velocity will be the fraction of total diplacement and time taken.

Range, $R=\frac{u^{2}sin2\theta }{g}$
Since $\theta ={45}^{\circ }$
$R=\frac{v^2}{g}$
Also $h=\frac{v^2}{4g}$
Hence net displacement, $\begin{array}{}D=\sqrt{(\frac{R}{2}{)}^2+h^2}=\frac{\sqrt{5}{v}^2}{4g}& \text{(3)}\end{array}$
Time taken to reach maximum height, $\begin{array}{}T=\frac{usin\theta }{g}=\frac{v}{\sqrt{2}g}& \text{(4)}\end{array}$
Dividing $\left(3\right)$ by $\left(4\right)$ for average velocity,
$V_{avg}=\frac{D}{T}=\frac{v}{2}\sqrt{\frac{5}{2}}$