It is a uniform circular motion. So the direction of velocity changes but not the magnitude. Therefore speed in considered constant. Again magnitude of acceleration does not change.

For freely falling body, velocity while the body, hit the ground \(v=\sqrt{2 g h}\)
\(v_{1}=\sqrt{2 g h_{1}} \text { and } v_{2}=\sqrt{2 g h_{2}} \)
\(\therefore \frac{m_{1} v_{1}}{m_{2} v_{2}}=\frac{m_{1} \sqrt{h_{1}}}{m_{2} \sqrt{h_{2}}}=\frac{m_{1}}{m_{2}} \sqrt{\frac{h_{1}}{h_{2}}}\)

Mass, length are scalars. Acceleration is a vector but magnitude of acceleration is a scalar

Unit of a (Planck's constant) - Js
Unit of c (Velocity of light) - ms^-1
Unit of G (Gravitational Constant) - \(\frac{\mathrm{Nm}^{2}}{\mathrm{Kg}^{2}}\)
\(\therefore \text { Unit of } \frac{h c}{G} \text { is }=\frac{J s \times m s^{-1}}{N m^{2} / k g^{2}} \)
\(=\frac{N m s \times m s^{-1} \times k g^{2}}{N m^{2}}[J=N m] =\mathrm{kg}^{2}\)

The relative velocity of the slow-moving train with respect to the boy is = (36 + 18) km h^-1
= 54 km h^-1 54 =\(\times \frac { 5 }{ 18 } \) ms^-1 =15 ms^-1
Since the boy will watch the full length of the other train, to find the time taken to watch the full train:
We have,
15=\(\frac { 90 }{ t } \) or t=\(\frac { 90 }{ 15 } \) = 6s

Degree is the unit of measurement which is used to determine the size of an angle.

(i) Barn,1 barn 10^-28m²
(ii) Acre,1 acre = 4047 m²
(iii) Hectare,1 hectare = 10⁴ m².

The distances of astronomical objects like stars, planets etc from the earth are huge. The distance on the earth are relatively small so it can be measured in km, m.
For Example:
The distance to be next nearest big galaxy Andromeda is  21,000,000,000,000,000,000 km.
i.e. 21\(\times\)10¹⁸ km.
This no is so large that it becomes hard to write and to interpret.
So astronomical units like the light year, parsec A.U for large distances.

Scalar is a property which can be described only by magnitude.
Examples:
Distance, mass, temperature, speed and energy.

No acceleration is towards the centre only in case of uniform circular motion.

Life of the man = 100 years
100 years includes 76 normal years and 24 leap years
Total no of days = 76\(\times\)365 + 24\(\times\)366 = 36524 days
Number of seconds = 36524\(\times\)24\(\times\)3600 = 3.155\(\times\)10° second
\(\text{Number of hearts beats}=\frac{Total\ no\ of\ seconds}{Time\ period\ of\ heart\ beat}=\frac{3.155\times10^9}{0.8s}=3.94\times10^9\)

Angle \(\theta\) = 1° 55'= 115' = (115\(\times\)60)"\(\times\) (4.85\(\times\) 10^-6) rad = 3.34\(\times\)10^-2 rad
Since 1" = 4.85\(\times\)10^-6 rad
Radius of the Earth = 6.4\(\times\)10⁶m
From the Figure AB is the diameter of the Earth (b) = 2\(\times\)6.4\(\times\)10⁶m
Distance of the Moon from the Earth x = ?
\(x={b\over \theta}={2\times 6.4 \times 10^6\over 3.34\times 10^{-2}}\)
x = 3.83\(\times\)10⁸m

Mass of a rocket, m = 7000 kg
Acceleration of the rocket, a = 30 m/s²
Speed of a rocket, v = 700 m/s
Force on the rocket, F = m(g + a)
F = m(g+a)   [\(\because\) g\(\rightarrow\)acceleration due to gravity is 9.8 ms^-2]
= 7000 (9.8 + 30)
= 7000(39.8)
F = 2.786\(\times\)10⁵N
Force, F v\(\frac { dm }{ dt } \)
\(\frac { dm }{ dt } =\frac { F }{ v } \)
= \(=\frac { 2.789\times { 10 }^{ 5 } }{ 700 } \)
Amount of gases ejected per second = 398 kgs^-1

Collinear vectors are those which act along the same line. The angle between them can be 0° or 180°.
(i) Parallel Vectors: If two vectors \(\overrightarrow { A } \)  and \(\overrightarrow { B } \) act in the same direction along the same line or on parallel lines, then the angle between them is 0°.

(ii) Anti-parallel vectors: Two vectors \(\overrightarrow { A } \ and\ \overrightarrow { B } \)  are said to be anti-parallel when they are in opposite directions along the same line or on parallel lines. Then the angle between them is 180°.

As soon as the monkey begins to fall, it will have downward vertical motion with acceleration due to gravity g.
Its equation of motion at any time t is given by
\(y_m=h-{1\over2}gt^2 \ \)  ...................(1)
When the bullet comes out of the gun, it has both vertical and horizontal components of velocity given by
\(v_{0x}=v_0 cos\theta;v_{\theta y}=v_o sin \theta \ \) ................(2)
Let us assume the horizontal distance between the monkey and hunter is 'd'.
At time t, the horizontal distance travelled by the bullet x = v_0x \(t=v_o cos \theta\)
When the horizontal position of bullet, x = d, the time d = v_oxT. It implies that T =\(d\over v_{ox}\)
At this time T, the vertical distance covered by the bullet is  
\(y_b=v_{oy}T-{1\over 2}gT^2={v_{0y}d\over v_{0x}}-{1\over2}gT^2\)
\(y_b={dv_o sin \theta \over v_o cos \theta}-{1\over2}gT^2=d\ tan \theta-{1\over 2}gT^2\) .................(3)
But from the figure we can write, tan \(\theta ={h\over d}\)
h = d tan \(\theta\)
By substituting this in the equation (3), we get
\(y=h-{1\over 2}gT^2\) ....................(4)
At this same time T, the vertical position of the monkey can be calculated from the equation (1)
\(y_m=h-{1\over 2}gT^2\) ..................(5)
Note that at the time T, the y coordinate of both monkey and bullet is same. It implies that the bullet will hit the monkey.

Integration is basically an area finding process. For certain geometric shapes we can directly find the area. But for irregular shapes the process of integration is used.

To find the area of the irregular shaped curve given by fix), we divide the area into rectangular strips as shown in the figure. The area under the curve is approximately equal to sum of areas of each rectangular strip.
This is given by \(A\approx f\left( a \right) \triangle x+f\left( { x }_{ 1 } \right) \triangle x+f\left( { x }_{ 2 } \right) \triangle x+f\left( b \right) \triangle x\)
Where f(a) is the value of the function f(x) at x = a,f(x₁) is the value of f(x) for x = x₁ and so on.
As we increase the number of strips, the area evaluated becomes more accurate. If the area under the curve is divided into N strips, the area under the curve is given by
\(A=\sum _{ n=1 }^{ N }{ { f }_{ n }\left( { x }_{ n } \right) \triangle { x }_{ n } } \)
As the number of strips goes to infinity,\(N\rightarrow \infty \) the sum becomes an integral,
\(A=\int _{ a }^{ b }{ f\left( x \right) dx } \)   (Note: As \(N\rightarrow \infty ,\triangle x\rightarrow 0\))
The integration will give the total area under the curve f(x).