Free JEE Advanced Test Practice Test - 11th and 12th 

(i) To find out time of flight here, we can analyze the motion in y-direction; we can use the formula $y=u_{y}t+\frac{1}{2}{a}_y{t}^2$. By analyzing motion in y-direction , the displacement of the particle in y-direction during motion is zero.

Now $u_y=usin\alpha =u.sin{37}^{\circ }=\frac{3}{5}\times 10=6\frac{m}{s}$

$a_y=-gcos\theta =-gcos{60}^{\circ }=-10\times \frac{1}{2}=-5\frac{m}{s^2}$

So, $y=u_{y}t+\frac{1}{2}{a}_y{t}^2\Rightarrow 0=6t-\frac{5}{2}{t}^2\Rightarrow t=\frac{12}{5}s$

(ii) To find out the distance traveled along AB, we have to analyze the motion in x-direction. So we have to use the formula

$x=u_{x}t+\frac{1}{2}{a}_x{t}^2$

Here $u_x=ucos\alpha =10cos{37}^{\circ }=10\times \frac{4}{5}=8\frac{m}{s}{a}_x=-gsin\theta =-10sin{60}^{\circ }=-10\times \frac{\sqrt{3}}{2}=-5\sqrt{3}\frac{m}{s^2}$

And $t=\frac{12}{5}s,x=8\times \frac{12}{5}-\frac{1}{2}.5\sqrt{3}{\left(\frac{12}{5}\right)}^2=\frac{96}{5}-\frac{5\sqrt{3}}{2}\times \frac{144}{25}=\frac{96}{5}-\frac{72\sqrt{3}}{5}=\frac{12}{5}(8-6\sqrt{3})m$

Initial velocity of 2 w.r.t. 1: $V_{2/1}=(v_{2}cos{\theta }_2-v_{1}cos{\theta }_1)\hat{i}+(v_{2}sin{\theta }_2-v_{1}sin{\theta }_1)\hat{j}$

Its $\hat{i}$ part is positive and $\hat{j}$ part is negative.

Trajectory of particle 2 will be a straight line OA w.r.t 1 as shown. Its slope is negative.

Line 1 - 2 will be parallel to OA at any time.

Relative acceleration between the two particles is zero. The distance between then at time $t$ is

$s$ = $\sqrt{(h-(v-vsin\theta t){)}^2+(vcos\theta t{)}^2}$

or $s^2$ = $(h-(v-vsin\theta t){)}^2$ + $(vcos\theta t{)}^{2}s$ is minimum when

or $\frac{d}{dt}\left(s^2\right)$ = $0$

or $2(h-(v-vsin\theta \left)t\right)(vsin\theta -v)+2v^{2}cos{}^2\theta t$ = $0$

or $t$ = $\frac{h}{2v}$

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 ${75}^{\circ }$. .

at t = 0, speed is 5 $\frac{m}{s}$, this means, u = 5 $\frac{m}{s}$. Also minimum speed is 3 $\frac{m}{s}$. For a projectile speed is minimum and equal to its horizontal component at the highest point of its trajectory, as the vertical component is zero at this point. Let $\theta$ be the angle of projection, then,
u cos $\theta$ = 3

$\Rightarrow$ 5 cos $\theta$ = 3

$\Rightarrow$ cos $\theta$ = $\frac{3}{5}$
$\theta$ = ${53}^{\circ }$
$\therefore$ Time of flight = $\frac{2usin\theta }{g}$ = $\frac{2\times 5\times sin{53}^{\circ }}{10}$
2$\times$5$\times$$\frac{4}{5}$$\times$ $\frac{1}{10}$
$\Rightarrow$ $\frac{4}{5}$s = 0.8s

$u_x$ = $25\hat{i}$ as the car is moving with a speed of 25 $\frac{m}{s}$ along +ve axis, given $\theta ={37}^{\circ }$

$\therefore$$\frac{u_y}{u_z}=tan{37}^{\circ }=\frac{3}{4}$

$\therefore$ the only possible option is $25\hat{i}+6\hat{j}+8\hat{k}$.

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)$ = $\frac{g{T}^2}{8}$

$\Rightarrow$ $T$ = $\sqrt{\frac{8(H-h)}{g}}$

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,

$2x$ = $\frac{R}{2}$

$\Rightarrow$ $x$ = $\frac{R}{4}$

$\Rightarrow$

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 $\theta$

$\Rightarrow$ $ucos\theta t$ = $\frac{3R}{4}$

$\Rightarrow$ $t$ = $\frac{3R}{4ucos\theta }$          ..........(1)

$h$ = $usin\theta t-\frac{1}{2}g{t}^2$           ..........(2)

substituting (I) in (II) we get

$h$ = $\frac{usin\theta \times 3R}{4ucos\theta }-\frac{1}{2}g\frac{9R^2}{16u^{2}co{s}^2\theta }$

we know $R$ in this case is $\frac{2u^{2}sin\theta cos\theta }{g}$

$\Rightarrow$ $h$ = $\frac{usin\theta X3X2u^{2}sin\theta cos\theta }{24ucos\theta g}-\frac{1}{2}g\frac{9\times {4u^4}^{2}si{n}^2\theta co{s}^2\theta }{g^{2}g\times {16}^4{u}^{2}co{s}^2\theta }$

$h$ = $\frac{3u^{2}si{n}^2\theta }{2g}-\frac{9u^{2}si{n}^2\theta }{8g}$

$h$ = $\frac{u^{2}si{n}^2\theta }{2g}[3-\frac{9}{4}]$

we know $\frac{u^{2}si{n}^2\theta }{2g}$ = $H$

$\Rightarrow$ $h$ = $\frac{3}{4}H$.

Method II

Equation of trajectory

$y$ = $xtan\theta -\frac{t^{2}tan\theta }{R}$

$y$ = $h$

$x$ = $\frac{3R}{4}$

$\Rightarrow$ $h=\frac{3R}{4}tan\theta -\frac{9R^{2}R}{16}\frac{tan\theta }{R}$

$h=\frac{3\times 2\times u^{2}sin\theta cos\theta sin\theta }{24\times gcos\theta }-\frac{9\times 2u^{2}sin\theta cos\theta sin\theta }{816\times u^2\times gcos\theta }$

$h$ = $\frac{3u^{2}si{n}^2\theta }{2g}-\frac{9u^{2}si{n}^2\theta }{8g}$

$h=\frac{u^{2}si{n}^2\theta }{2g}(3-\frac{9}{4})$

$h$ = $\frac{3}{4}H$.

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 $(228-36)$

= $192ft$.

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.

Let in 2 sec body reaches upto point A and after one more sec upto point B.

Total time of ascent for a body is given 3 sec i.e. $t=\frac{usin\theta }{g}=3$

$\therefore usin\theta =10\times 3=30.............\left(i\right)$

Horizontal component of velocity remains always constant

$ucos\theta =vcos{30}^{\circ }..........\left(ii\right)$

For vertical upward motion between point O and A

$vsin{30}^{\circ }=usin\theta -g\times 2[Usingv=u-gtvsin{30}^{\circ }=30-20[Asusin\theta =30]$

Substituting this value in equation (ii) u coa $\theta =20cos{30}^{\circ }=10\sqrt{3}......\left(iii\right)$

From equation (i) and (iii) $u=20\sqrt{3}and\theta ={60}^{\circ }$

At point N angle of projection of the body will be ${45}^{\circ }$. Let velocity of projection at this point is v. If the body just manages to cross the well then

Range = Diameter of well

$\frac{v^{2}sin2\theta }{g}=40[As\theta ={45}^{\circ }]v^2=400\Rightarrow v=20m/s$

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) = $-gsin\alpha =-gsin{45}^{\circ }$, distance of inclined plane (s) = $20\sqrt{2}m$

$(20{)}^2=u^2-2\frac{g}{\sqrt{2}}.20\sqrt{2}[Using{v}^2=u^2+2as]u^2={20}^2+400\Rightarrow u=20\sqrt{2}m/s.$

When body projected with initial velocity $\vec{u}$ by making angle $\alpha$ with the horizontal. Then after time t, (at point P) it’s direction is perpendicular to $\vec{u}$

Magnitude of velocity at point P is given by v = $ucot\alpha$ (from sample problem no. 9)

For vertical motion : Initial velocity (at point O) $=usin\alpha$

Final velocity (at point P) $=-vcos\alpha =-ucot\alpha cos\alpha$

Time of flight (from point O to P) = t

Applying first equation of motion $v=u-gt$

$-ucot\alpha cos\alpha =usin\alpha -gt\therefore t=\frac{usin\alpha +ucot\alpha cos\alpha }{g}=\frac{u}{gsin\alpha }[si{n}^2\alpha +co{s}^2\alpha ]=\frac{ucosec\alpha }{g}$

Formula for calculation of time to reach the body on the ground from the tower of height ‘h’ (If it is thrown vertically up with velocity u) is given by $t=\frac{u}{g}[1+\sqrt{1+\frac{2gh}{u^2}}]$

So we can resolve the given velocity in vertical direction and can apply the above formula.

Initial vertical component of velocity $usin\theta =50sin30=25m/s$

$\therefore t=\frac{25}{9.8}[1+\sqrt{1+\frac{2\times 9.8\times 70}{(25{)}^2}}]=7.11sec$