Solution:

If the speed of the particle in circular path remains constant, the motion is uniform circular motion. We know, the arc length `x` covered with an angular displacement `theta` is `x = r theta`.
Differentiating,

` (dx)/(dt) = r (d theta)/(dt) \ \ ∵ r` is constant, `:. v = r omega`.

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Solution:

The resultant vector of two or more vectors is defined as that single vector which produces the same effect as is produced by individual vectors together. If `F_1` and `F_ 2` are the two forces the magnitude of the resultant `F` is given by,

`F = sqrt ( F_1^2 + F_2^2 + 2 F_1 F_2 cos theta)`

The angle `theta` between them is,

` theta = cos^(-1) ( (F^2 - F_1^2 - F_2^2)/(2 F_1 F_2) )`

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Solution:

Dot product of two vectors `vec a` and `vec b` is defined
as, `vec a . vec b = | vec a | | vec b | cos theta` where `theta` is the angle
between `vec a` and `vec b`.

Let `vec (O A) = vec a , vec (O B) = vec b`

and `angle A O B = theta`

Then, `OT = | vec b | cos theta`

`vec a . vec b = | vec a | | vec b | cos theta`

` = OA xx OT`

`vec a . vec b = OA`. Projection of `vec b` on `vec a` .

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Solution:

The coordinates of one projectile as seen from another projectile are :

`X= x_1 - x_2 = (u_1 cos theta_1 - u_2 cos theta_2 ) t`

`Y =y_1 - y_2` ,

`= ( u_1 sin theta_1) t - 1/2 g t^2 - ( u_2 sin theta_2) t + 1/2 g t^2`

`= (u_1 sin theta_1 - u_2 sin theta_2 ) t`

`:. Y/X = ( (u_1 sin theta_1 - u_2 sin theta_2)t)/( (u_1 cos theta_1 - u_2 cos theta_2)t)`

` = ( u_1 sin theta_1 - u_2 sin theta_2)/( u_1 cos theta_1 - u_2 cos theta_2)`

` = m ` (constant)

or, `Y = m X`

This equation represents straight line. Hence proved.

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Solution:

When a body is projected in the air in any direction, then the body is called a projectile.
(a) Suppose a body is projected with velocity `u` at an angle `theta` with the horizontal, `P(x, y)` is any point on its trajectory at time `t`.
Horizontal component of velocity is unaffected by gravity, but the vertical component `(u sin theta)` changes due to gravity

`:. x = ( u cos theta) t`.

` y = ( u sin theta ) t - 1/2 g t^2`

` = u sin theta xx x/(u cos theta) - 1/2 g ( x/(u cos theta))^2`

` y = x tan theta - (gx^2)/(2 u^2 cos^2 theta)` ............(i)

It represents the equation of a parabola, hence the path followed by a projectile is a parabola.
(b) The greatest vertical distance attained by the projectile above the horizontal plane from the point of projection is called maximum height.
Maximum height, `LN = H`

(i) At maximum height
`v = 0`
`:. u^2 - u_y^2 = -2 \ \ g H`, where
`u_y = u sin theta`
or `(u sin theta)^2 = 2 \ \ g H`
or `H = (u^2 sin^2 theta)/(2g)`

(ii) At maximum height
`v = 0`
`:. 0 = u sin theta - g t`
or ` t = ( u sin theta)/g`

But time of flight,

`T = 2t = (2 u sin theta)/g`

(iii) When the body returns to the same horizontal level `y = 0`

`:. 0 = x tan theta - (gx^2)/( 2 u^2 cos^2 theta)`

or ` x tan theta = (gx^2)/( 2 u^2 cos^2 theta)`

or ` x = ( 2 u^2 sin theta cos theta)/g = (u^2 sin 2 theta)/g`

But coordinates of `M` are `(R, 0)`. Putting `x = R`, we have

` R = ( u^2 sin 2 theta)/g`

(c) ` theta = 45^o`

(d) When an object is projected with velocity `u` making an angle `theta` with horizontal direction.

` R_1 = ( u^2 sin 2 theta)/g` ...........(i)

When an object is projected with u making an angle `(90° - theta)`

`R_2 = ( u^2 sin 2 (90^o - theta))/g`

` = u^2/g sin (180° - 2 theta)`

`= u^2/g sin 2 theta` ..............(ii)

from (i) and (ii) `R_1 = R_2`

`:.` The horizontal range is same for two complementary angles.

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Solution:

Consider a body in a circular path of radius r, with a speed v. The velocity direction is tangential at any point in the path. The position vectors at A and B are represented by two sides of an isosceles triangle first.
The change in position vector is indicated by `bar(AB) = Delta r`. The velocity at A and B are along the tangents at these points and the change in velocity will complete an isosceles triangle of velocities.

`bar MN = bar Delta v` Since the triangles are similar,

`(Delta v)/(Delta r) = v/r => Delta v = v/r . Delta r`

`Lt_(Delta t -> 0) (Delta v)/(Delta t) = Lt_(Delta t -> 0) v/r (Delta r)/(Delta t)`

`:. (dv)/(dt) = v/r . v => a = v^2/r`

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Solution:

Two vectors having the same initial point are called co-initial vectors. Two vectors which either act along the same line or along parallel lines are called collinear vectors.

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Solution:

Consider a projectile projected at an angle `theta` to the horizontal with velocity `u`. The horizontal and vertical components initially with velocity are `u cos theta` and `u sin theta` respectively. Vertical velocity at highest point is zero, due to acceleration due to gravity acting vertically downwards.
Using ` v = u + a t` we have,
`0 = u sin theta - g t`

` => t = (u sin theta)/g`

The time to reach topmost point `t = (u sin theta)/g`

Using `v^2 = u^2 + 2 a s`, we have

`0 = u^2 sin^2 theta - 2 g \ \ h_(m a x)`

` h_(m a x) = ( u^2 sin^2 theta)/(2 g)`

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Solution:

The required angle of projection for max. horizontal range is `45°`.

`R_(max) = u^2/g`

Max. height,

` H = (u^2 sin^2 theta)/(2g)`

` = ( u^2 sin^2 45^o)/(2g) = u^2/(4g)`

` (R_(max))/H = (u^2//g)/(u^2//4g) = 4`

or ` R_(max) = 4 H`

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Solution:

Parallelogram law of vector addition. If two vectors `vec A` and `vec B` represent two adjacent sides of a parallelogram, the sum of the vectors is represented by the diagonal of the parallelogram.

Let `vec A` and `vec B` be two vectors at an angle `theta` between them. According to the law of parallelogram of vectors, the diagonal of the parallelogram indicates the sum of the other two sides/vectors `vec A` and `vec B`.

`| vec (OQ) | = | vec A + vec B | = sqrt ( O T^2 + T Q^2)`

`| vec R | = sqrt ( (A + B + cos theta)^2 + ( B sin theta)^2)`

` R = sqrt ( A^2 + B^2 + 2 AB cos theta) `

The resultant `R` is at an angle `alpha` to `vec A` given by,

` alpha = tan^(-1) ( (B sin theta)/(A + B cos theta) )`

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Solution:

At `45^o`, the projectile has maximum range.

At `(45° - theta ) ` the range is,

` R_1 = (u^2 sin [ 2 ( 45^o - theta) ])/g`

` = ( u^2 sin (90^o - 2 theta) )/g`

` = (u^2 cos 2 theta) /g`

At `(45° + theta)` the range is,

` R^2 = ( u^2 sin [ 2 (45^o + theta) ] )/g`

` = (u^2 sin (90^o + 2 theta) )/g`

` = (u^2 cos 2 theta) /g`

` R_1 = R_2`

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