Answer: Option C. -> 3H4
:
C
The ball will cover the same distance as only velocity's direction is changed magnitude is the same. So the same distance will be covered in opposite direction.


Now from the figure it's clear,
$2x$ = $\frac{R}{2}$
$⇒$ $x$ = $\frac{R}{4}$
$⇒$
So it means I have to find the height of the ball when it has travelled $\frac{3R}{4}$ along the x-axis.
Method I
Assume initial speed of projection is $u$ and angle $\mathrm{θ}$
$⇒$ $ucos\mathrm{θ}t$ = $\frac{3R}{4}$
$⇒$ $t$ = $\frac{3R}{4ucos\mathrm{θ}}$ ..........(1)
$h$ = $usin\mathrm{θ}t−\frac{1}{2}g{t}^2$ ..........(2)
substituting (I) in (II) we get
$h$ = $\frac{usin\mathrm{θ}×3R}{4ucos\mathrm{θ}}−\frac{1}{2}g\frac{9R^2}{16u^{2}co{s}^2\mathrm{θ}}$
we know $R$in this case is $\frac{2u^{2}sin\mathrm{θ}cos\mathrm{θ}}{g}$
$⇒$ $h$ = $\frac{usin\mathrm{θ}X3X2u^{2}sin\mathrm{θ}cos\mathrm{θ}}{24ucos\mathrm{θ}g}−\frac{1}{2}g\frac{9×{4u^4}^{2}si{n}^2\mathrm{θ}co{s}^2\mathrm{θ}}{g^{2}g×{16}^4{u}^{2}co{s}^2\mathrm{θ}}$
$h$ = $\frac{3u^{2}si{n}^2\mathrm{θ}}{2g}−\frac{9u^{2}si{n}^2\mathrm{θ}}{8g}$
$h$ = $\frac{u^{2}si{n}^2\mathrm{θ}}{2g}[3−\frac{9}{4}]$
we know $\frac{u^{2}si{n}^2\mathrm{θ}}{2g}$ = $H$
$⇒$ $h$ = $\frac{3}{4}H$.
Method II
Equation of trajectory
$y$ = $xtan\mathrm{θ}−\frac{t^{2}tan\mathrm{θ}}{R}$
$y$ = $h$
$x$ = $\frac{3R}{4}$
$⇒$ $h=\frac{3R}{4}tan\mathrm{θ}−\frac{9R^{2}R}{16}\frac{tan\mathrm{θ}}{R}$
$h=\frac{3×2×u^{2}sin\mathrm{θ}cos\mathrm{θ}sin\mathrm{θ}}{24×gcos\mathrm{θ}}−\frac{9×2u^{2}sin\mathrm{θ}cos\mathrm{θ}sin\mathrm{θ}}{816×u^2×gcos\mathrm{θ}}$
$h$ = $\frac{3u^{2}si{n}^2\mathrm{θ}}{2g}−\frac{9u^{2}si{n}^2\mathrm{θ}}{8g}$
$h=\frac{u^{2}si{n}^2\mathrm{θ}}{2g}(3−\frac{9}{4})$
$h$ = $\frac{3}{4}H$.

Answer: Option D. -> 20 √2 ms−1
:
D
At point N angle of projection of the body will be ${45}^{∘}$. Let velocity of projection at this point is v. If the body just manages to cross the well then
Range = Diameter of well
$\frac{v^{2}sin2\mathrm{θ}}{g}=40[As\mathrm{θ}={45}^{∘}]v^2=400⇒v=20m/s$
But we have to calculate the velocity (u) of the body at point M.
For motion along the inclined plane (from M to N)
Final velocity (v) = 20 m/s, acceleration (a) = $−gsin\mathrm{Î\pm }=−gsin{45}^{∘}$, distance of inclined plane (s) = $20\sqrt{2}m$
$(20{)}^2=u^2−2\frac{g}{\sqrt{2}}.20\sqrt{2}[Using{v}^2=u^2+2as]u^2={20}^2+400⇒u=20\sqrt{2}m/s.$

Answer: Option B. -> 49.0 N–s
:
B
Change in momentum between complete projectile motion = $2musin\theta =2\times 0.5\times 98\times sin{30}^{\circ }=49N-s$

Answer: Option C. -> u sin θ
:
C
Since horizontal component of velocity remain always constant therefore only vertical component of velocity changes.
Initially vertical component $usin\theta$
Finally it becomes zero. So change in velocity $=usin\theta$

Answer: Option D. -> 1 kg m sec−1
:
D
Horizontal momentum remains always constant. So change in vertical momentum $\left(\Delta \overrightarrow{p}\right)$ = Final vertical momentum – Initial vertical momentum $=-musin\theta$
$|\Delta P|=0.1\times 20\times sin{30}^{\circ }=1kgm/sec$

Answer: Option B. -> mv3(4√2g)
:
B
$L=\frac{m{u}^{3}cos\theta si{n}^2\theta }{2g}=\frac{m{v}^3}{\left(4\sqrt{2}g\right)}[As\theta ={45}^{\circ }]$

Answer: Option A. -> 605.3 m
:
A
The horizontal distance covered by bomb,
$BC=v_H×\sqrt{\frac{2h}{g}}=150\sqrt{\frac{2×80}{10}}=600m$

$∴$ The distance of target from dropping point
$∴$ The distance of target from dropping point of bomb,
$AC$ = $\sqrt{A{B}^2+B{C}^2}$ = $\sqrt{(80{)}^2+(600{)}^2}$ = $605.3m$

Answer: Option B. -> 6R
:
B
$R=\frac{u^{2}sin2\theta }{g}\therefore R\propto \frac{1}{g}\frac{R_{Moon}}{R_{Earth}}=\frac{g_{Earth}}{g_{Moon}}=6[\because g_{Moon}=\frac{1}{6}{g}_{Earth}]\therefore R_{Moon}=6R_{Earth}=6R$

Answer: Option C. -> A parabola
:
C
The path will be parabolic

Answer: Option A. -> (Zero)
:
A
Both masses will collide at the highest point of their trajectory with equal and opposite momentum. So net momentum of the system will be zero.