Velocity of aeroplane in horizontal direction is
v_ox = 600 km/hr = 600 x \({5\over18}ms^{-1}\)
\(={500\over 3}ms^{-1}\)
This velocity remains constant throughout the flight of the body.
v_oy = 0 and y = h = 1960 m 
Let t = the time taken by the body to reach the ground
Now y = v_oy  t + \({1\over2}gt^{2}\)
Here, y = h = 1960 m; v_oy = 0 (initial vertical velocity)
\(\therefore \ 1960={1\over2}\times 9.8t^2\)
\(\therefore t=\sqrt{1960\over4.9}s=\sqrt{400}s=20s\)
Distance travelled by the body in the horizontal direction
\(=v_{ox}t={500\over3}\times20={10000\over3}=3333m\)
= 3.333 km
\(\therefore\) AB = 3.333 km

Here m = 200 g = 0.2 kg
r = 80 cm = 0.8 m
V = 1.5 ms^-1
\(\theta\) = 60°
\(\therefore\) Required tension in the string
\(T={mV^2\over r}+mg cos \theta\)
\(={0.2\times(1.5)^2\over 0.8}+0.2\times 10\times cos 60^o\)
\(={2.25\over 4}+0.2\times 10 \times {1\over2}\)
= 0.56 + 1
= 1.56 N

Here m= 20 kg, F = - 50 N (retardation force)
As F = ma
\(\Rightarrow a={F\over m}={-50\over 20}=-2.5 ms^{-2}\)
Using equation, v = u + at
Given, u 15 ms^-1, v = 0
Now, 0 15 + (- 25) t
or, t = 6 s.

Given, initial velocity (u) = 40m/s
Height of the hall (H) = 25m
Let the angle of projection of the ball be \(\theta \), when maximum height attained by it be 25m.
Maximum height attained by the ball
\(H=\frac { { u }^{ 2 }{ sin }^{ 2 }\theta }{ 2g } \Rightarrow 25=\frac { { (40) }^{ 2 }{ sin }^{ 2 }\theta }{ 2\times 9.8 }\)
\(or\ { \sin }^{ 2 }\theta =\frac { 25\times 2\times 9.8 }{ 1600 } =0.3068\)
\(or\ \sin\theta =0.5534=\sin{ 33.6 }^{ 0 }\)
\(or\ \theta ={ 33.6 }^{ 0 }\)
\( \therefore \text{ Horizontal range (R)}=\frac { { u }^{ 2 }\sin2\theta }{ g }\)
\( \\ =\frac { { (40) }^{ 2 }sin2\times { 33.6 }^{ 0 } }{ 9.8 } =\frac { 1600\times sin{ 67.2 }^{ 0 } }{ 9.8 } \)
\(\\ =\frac { 1600\times 0.9219 }{ 9.8 } =150.5m\)

As the cyclist leans by an angle θ with the vertical, the normal,can be acting as shown. If v is the speed of the cyclist, the necessary centripetal force is provided by the component N sin θ of normal reaction and N cos θ provides balancing for weight.

ஃ N sin θ = mv²/r, N cos θ = mg
Dividing \(\tan \theta=\frac{v^{2}}{r g}\)
\(\Rightarrow \ \theta=\tan ^{-1}\left(\frac{v^{2}}{r g}\right)\) 

Magnitude is same when θ between \(\overrightarrow{\mathbf{A}}\) and \(\overrightarrow{\mathbf{B}}\) is 90°, and is different if θ is other than 90°, but the direction is opposite between them.

(i) The part correctly describes the trajectory of the stone after the spring breaks because when a stone tied to one end of a string is whirled round in a circle then velocity of the stone at any point is along the tangent at that point. If the string breaks suddenly, then stone flies off tangentially, along the direction of its velocity.
(ii) The part correctly describes the trajectory of the stone after the spring breaks because when a stone tied to one end of a string is whirled round in a circle then velocity of the stone at any point is along the tangent at that point. If the string breaks suddenly, then stone flies off tangentially, along the direction of its velocity.
(iii) The part correctly describes the trajectory of the stone after the spring breaks because when a stone tied to one end of a string is whirled round in a circle then velocity of the stone at any point is along the tangent at that point. If the string breaks suddenly, then stone flies off tangentially, along the direction of its velocity.

Here, r = 1 km = 1000 m,
v = 900 kmh^-1 = 900 x (1000m(/(60 x 60s) = 250 ms^-1
Centripetal acceleration, a = \(\frac { { v }^{ 2 } }{ r } =\frac { { (250) }^{ 2 } }{ 1000 } \)
Now,  \(\frac { a }{ g } =\frac { { (250) }^{ 2 } }{ 1000 } \times\frac { 1 }{ 9.8 } =6.38\)

The situation is sketched in figure. We assign a letter to each body in relative motion, P passenger, S sidewalk, G ground. The relative velocities V_PS and V_SG are given
v_PS = 5.00 km /h, to the right
v_SG = 3.00 km /h, to the right

(i) Here we must find the magnitude of the vector v_PG given the magnitude and direction of two other vectors. We find the velocity v_PGby using the relation
V_PG = V_PS +V_SG
Here the vectors are parallel, and so the vector addition is quite simple see figure. We add vectors by adding magnitudes
V_PG = V_PS +V_SG
= 5.00 km / h + 3.00 km/h
= 8.00km/h
(ii) The length of the sidewalk is 135 m, and so this is the distance \(\triangle\)x_G the passenger travels relative to the ground. So, our problem is to find \(\triangle\)t when \(\triangle\)x_G = 135 m. The rate at which this distance along the ground is covered by the passenger is v _PG where
\(v_{P G}=\frac{\Delta x_{G}}{\Delta t}\)
Therefore,
\(\Delta t=\frac{\Delta x_{G}}{v_{P G}}=\frac{135 \mathrm{~m}}{8.00 \mathrm{~km} / \mathrm{h}\left(\frac{1.00 \mathrm{~m} / \mathrm{s}}{3.60 \mathrm{~km} / \mathrm{h}}\right)}=60.8 \mathrm{~s}\)
The problem here is to determine how much of the sidewalk's surface the passenger moves over. If he was standing still and not walking along the surface, he would cover none of it. Because he is moving relative to the surface at velocity v_PS he does move some distance \(\triangle\)x _s relative to the surface. The problem is to find \(\triangle\)s_x , when \(\triangle\)t = 60.8 s, since we found in part (b) that this is the time interval during which he is on the moving sidewalk. His velocity relative to the sidewalk is \(v_{P S}=\Delta x_{\mathrm{s}} / \Delta t\) and so
\(\Delta x_{s}=v_{P S} \Delta t\)
\(=(5.00 \mathrm{~km} / \mathrm{h})\left(\frac{1.00 \mathrm{~m} / \mathrm{s}}{3.60 \mathrm{~km} / \mathrm{h}}\right)(60.8 \mathrm{~s})\)
= 84.4m

Here, m₁ = 4 kg, m₂ = 5 kg
Force applied on block P = 12 N
This force must be equal to the kinetic .friction applied on P by Q.
\(12={ F }_{ k }={ \mu }_{ k }R={ \mu }_{ k }{ m }_{ 1 }g \)
\(\\ 12={ \mu }_{ k }\times 4g\ or\ { \mu }_{ k }={ \frac { 12 }{ 4g } }=\frac { 3 }{ g }\)
acceleration of P=\(\frac { 12 }{ 4 } =3m/s\)
 
When force F is applied on Q to create common motion in P & Q, the forces created (friction) is shown in the diagram above. F, force will be created on P and F₂ force will be created on Q.
Considering forces on Q, we have
 \(F-{ F }_{ 2 }(net\quad force)={ m }_{ 2 }\times a={ m }_{ 2 }\times 3\)
Since,
 \({ F }_{ 1 }={ F }_{ 2 }={ \mu }_{ 1 }{ m }_{ 1 }g\) 
\(\\ F-{ \mu }_{ k }{ m }_{ 2 }g={ m }_{ 2 }\times 3\)
\(\\ F={ \mu }_{ k }{ m }_{ 2 }g+{ m }_{ 2 }\times 3=\frac { 3 }{ g } \times 4\times g+5\times 3=12+15=27N\)
As this force moves both the blocks together on a smooth table, so the acceleration produced is
\(a=\frac { F }{ { m }_{ 1 }+{ m }_{ 2 } } =\frac { 27 }{ 4+5 } =3m/s\)