Answer: Option B. -> θ=tan−1(12cot α)
:
B

As the ballhas to hit the inclined plane normally, so in that position the x-component of velocity will be zero and velocity will have y-component only.
The ball will hit the incline normally if its parallel component of velocity reduces to zero during the time of flight.
By analyzing this motion along incline, i.e., x-direction $v_x=u_x+a_{x}t$
Here $v_x=0,u_x=v_{0}cos\theta ,a_x=-gsin\alpha$
$0=v_{o}cos\theta -\left(gsin\alpha \right)T\Rightarrow T=\frac{v_{o}cos\theta }{gsin\alpha }$ ........(i)
Also the displacement of the particle in y-direction will be zero. Using
$y=u_{y}t+\frac{1}{2}{a}_y{t}^2\Rightarrow 0=v_{o}sin\theta .T-\frac{1}{2}gcos\alpha .T^2$
This gives T = $\frac{2v_{o}sin\theta }{gcos\alpha }$ ...............(ii)
From (i) and (ii), we have $\frac{v_{o}cos\theta }{gsin\alpha }=\frac{2v_{o}sin\theta }{gcos\alpha }\Rightarrow \frac{cos\theta }{sin\alpha }=\frac{2sin\theta }{cos\alpha }\Rightarrow 2tan\theta tan\alpha =1\Rightarrow tan\theta =\left[\frac{1}{2}cos\alpha \right]$
$\Rightarrow =ta{n}^{-1}\left(\frac{1}{2}cot\alpha \right)$ which is the required angle of projection.

Answer: Option D. -> 9 sec and 1800 m
:
D

$Timeofdescentt=\sqrt{\frac{2h}{g}}=\sqrt{\frac{2\times 396.9}{9.8}}\Rightarrow t=9secandhorizontaldistanceS=u\times tS=\left(\frac{720\times 5}{18}\right)\times 9=1800m$

Answer: Option C. -> t = √8g(H−h)
:
C

Time of flight for this part of motion (between points A and B, both at height h) assuming speed of projection to be $v$ and angle of projection to be $\alpha$
$T=\frac{2vsin\alpha }{g}$ ---------(I)
Now we have to eliminate $v$ & $\alpha$ as these variables we assumed.
Hmmmmm..........How do we do that, okay what else is given?
Yes, the maximum height i.e. $(H-h)$.
$\Rightarrow$ $(H-h)$ = $\frac{v^{2}si{n}^2\alpha }{2g}$ ----------(II)
Thank goodness!
Now from equation (I) $\Rightarrow$ $vsin\alpha$ = $\frac{Tg}{2}$
Equation (II) $\Rightarrow$ $(H-h)$ = $\frac{T^2{g}^2}{4\left(2g\right)}$
$\Rightarrow$$(H-h)$ = $\frac{g{T}^2}{8}$
$\Rightarrow$ $T$ = $\sqrt{\frac{8(H-h)}{g}}$

Answer: Option A. -> 15∘,75∘
:
A

Let AB be the boat, to touch A, Range should be 5m.
u= 10 $\frac{m}{s}$, g = 10m/$s^2$, R = 5 m,
$\therefore$R = $\frac{u^{2}sin2\theta }{g}$ $\Rightarrow$ 5 = $\frac{{10}^{2}sin2\theta }{10}$$\Rightarrow$ sin 2$\theta$ = $\frac{1}{2}$
$\therefore$2$\theta$ = ${30}^{\circ }$,15${\theta }^{\circ }$(as in$\pi$-$\theta$=sin$\theta$)
$\therefore$$\theta$ = ${15}^{\circ }$,${75}^{\circ }$
To throw at B
R = 6m
$\therefore$ 6 = $\frac{{10}^{2}sin2\theta }{10}$$\Rightarrow$sin 2$\theta$ = $\frac{3}{5}$
$\Rightarrow$2$\theta$ = ${37}^{\circ }$,${143}^{\circ }$(sin($\pi$-$\theta$)= sin $\theta$)
$\Rightarrow$$\theta$ = ${18.5}^{\circ }$,${71.5}^{\circ }$
$\therefore$Minimum is ${15}^{\circ }$ and maximum is ${75}^{\circ }$
$\therefore$For a successful shot,
${15}^{\circ }$$\le$$\theta$$\le$${18.5}^{\circ }$ and ${71.5}^{\circ }$$\le$$\theta$$\le$${75}^{\circ }$
$\therefore$Minimum is ${15}^{\circ }$ and maximum is ${75}^{\circ }$. .

Answer: Option D. -> 192 feet
:
D

In the given figure we can find $\theta$ by using trigonometry.
$tan\theta$ = $\frac{171}{228}$ = $\frac{3}{4}$
$\Rightarrow$ $\theta$ = ${37}^{\circ }$.
This means the ball was thrown at an angle ${37}^{\circ }$ to horizontal also its given that it was thrown with velocity 15 ft/s.
The dotted line shows how the trajectory of the ball thrown
So we know 2 D is nothing but 2 one dimensional motion.
Let's break the components:
$u_y$ = $-15sin{37}^{\circ }$ = $-15\times \frac{3}{5}$ = $-9ft/s$
$a_y$ = $-32ft/s^2$
$s_y$ = $-171ft$.
$\Rightarrow$ $s_y$ = $u_{y}t+\frac{1}{2}{a}_y{t}^2$
$-171$ = $-9t-\frac{32}{2}{t}^2$
$16t^2+9t-171=0$
$16t^2+57t-48t-171=0$
$\Rightarrow$ $t$ = $3sec$
We need to know in that time the projectile covered how much of horizontal distance.
So $u_x=15cos{37}^{\circ }$ = $15\times \frac{4}{5}$ = $12ft/s$
$a_x$ = $0$; $t$ = $3$
$s_x$ = $u_{x}t$ = $12\times 3$ = $36ft$
$\Rightarrow$ The packet went only 36 ft towards his friend
$\Rightarrow$ The packet fell short by $(228-36)$
= $192ft$.