Angle between velocity vector and acceleration vector in uniform circular motion is 90°.

To prove\(\left| \overset { \rightarrow }{ A } +\overset { \rightarrow }{ B } \right| \ge \left| \left| \overset { \rightarrow }{ A } \right| -\left| \overset { \rightarrow }{ B } \right| \right| \)
From \(\Delta \)OPS, we have OS + Ps > OP or OS > lOP - Ps I or OS > lOP _ OQ I          ...(ii)  (\(\therefore\)PS = OQ)
The modulus of (OP - PS) has been taken because the L.H.S. is always positive but the R.H.5. may be negative if OP < PS. Thus from
(iii) we have.\(\left| \overset { \rightarrow }{ A } +\overset { \rightarrow }{ B } \right| >\left| \left| \overset { \rightarrow }{ A } \right| -\left| \overset { \rightarrow }{ B } \right| \right| \)          ...(iv)
If the two vectors A and 13are acting along a straight line in opposite directions, then
\(\left| \overset { \rightarrow }{ A } +\overset { \rightarrow }{ B } \right| =\left| \left| \overset { \rightarrow }{ A } \right| -\left| \overset { \rightarrow }{ B } \right| \right| \)         ...(v)
Combining the conditions mentioned in (iv) and (v) we get.
\(\left| \overset { \rightarrow }{ A } +\overset { \rightarrow }{ B } \right| \ge \left| \left| \overset { \rightarrow }{ A } \right| -\left| \overset { \rightarrow }{ B } \right| \right| \) 

If three vectors acting on a point object at the same time are represented in magnitude and direction by the three sides of a triangle Q taken in the same order, their resultant is zero. The object is said to be in equilibrium.
 

An object that is in flight after being projected is called a projectile.

The times of flights are \( { T }_{ 1 }=\frac { 2u\sin { \theta } }{ g }\) and
\({ T }_{ 2 }=\frac { 2u\sin { \left( \frac { \pi }{ 2 } -\theta \right) } }{ g } =\frac { 2u\cos { \theta } }{ g }\)
\(\therefore \frac { T_{ 1 } }{ { T }_{ 2 } } =\frac { \sin { \theta } }{ \cos { \theta } } =\tan { \theta }\)

No, it is not a projectile, because a projectile should have two component velocities in two mutually perpendicular directions but in this case, the body has velocity only in one direction while going up or coming down.

Given, A = 20 N, B = 20 N, R = 30 N, \(\theta \) = ?
 \(R=\sqrt { { A }^{ 2 }+{ B }^{ 2 }+2AB\cos { \theta } }\)
\( \\ 30=\sqrt { { 2 }0^{ 2 }+{ 2 }0^{ 2 }+2\times 20\times 20\cos { \theta } } \)
\(\\ \cos { \theta } =\frac { { 30 }^{ 2 }-{ 2 }0^{ 2 }-{ 2 }0^{ 2 } }{ 2\times { 2 }0^{ 2 } } =0.125\)
\( \Rightarrow \quad \theta ={ 82 }^{ 0 }{ 49 }^{ ' }\) 

74 m

\({ A }_{ x }=17.7km,{ A }_{ y }=-17.7km,B_{ x }=20.0km,B_{ y }=34.6km\)

No,adding any two vectors is not meaningful because only vectors of the same dimensions i.e. having same unit can be added.

v = 8km/h, θ = 53⁰7'33''

Here, actual path length travelled, s = 23 km; Displacement = 10 km;
Time taken, t  = 28 min = \(\frac{28}{60} h\)
(a) Average speed of taxi = \(\frac{\text { actual path length }}{\text { time taken }}=\frac{23}{\frac{28}{60}} k \frac{m}{h}=49.3 \mathrm{~km} / \mathrm{h}\)
(b) Magnitude of average velocity = \(=\frac{\text { displacement }}{\text { time taken }}=\frac{10}{\frac{28}{60}} \mathrm{~km} / \mathrm{h}=21.4 \mathrm{~km} / \mathrm{h}\)
The average speed is not equal to the magnitude of average velocity. The two are equal for the motion of taxi along a straight path in one direction.

When a stone is going around a circular path, the instantaneous velocity of stone is acting as tangent to the circle. When the string breaks, the centripetal force stops to act. Due to inertia, the stone continues to move along the tangent to circular path. So, the stone flies off tangentially to the circular path.

The sum of two vectors is maximum, when both the Vectors are in the same direction and is minimum when they act in opposite direction.
As, \(R =\sqrt { A^{ 2 }+B^{ 2 }+2ABcos\theta }\)
(i) For R to be maximum, cos θ = +1
\(R_{ max }\sqrt { A^{ 2 }+B^{ 2 }+2AB } =A+B\)
(ii) For R to be minimum
cos θ   = -1 or θ  = 180⁰
\(R_{ min }\sqrt { A^{ 2 }+B^{ 2 }+2AB\left( -1 \right) } =A-B\)

Two only, provided that they are acting in opposite directions.

The two vectors are parallel and acting in the same direction i.e.θ = 0⁰

Yes, because they cannot be represented by the three sides of a triangle taken in the same order

Maximum height = b =\(u^2sin^2\theta\over 2g\)
or \(sin^2 \theta={2bg\over u^2}\)
Horizontal range, a =\({u^2sin2\theta\over 2g }={u^2sin\theta cos \theta\over g }\)
\(\Rightarrow 2 sin \theta cos \theta={ag\over u^2}\)
or \(4 sin^2 \theta cos^2 \theta={a^2g^2\over u^4}\)
\(\Rightarrow 4 sin^2 \theta (1-sin^2 \theta)={a^2g^2\over u^4}\) ............(i)
or \( 4({2bg\over u^2})[{1-{2bg\over u^2}}]={a^2g^2\over u^4}\)
or  \({8bg\over u^2}-{16b^2g^2\over u^4}={a^2g^2\over u^4}\)
\(\Rightarrow a^2g^2+16b^2g^2=u^28bg\)
or \(u^2={a^2g^2+16b^2g^2\over 8bg}\)
or \(u=[2g(b+{a^2\over 16b})]^{1/2}\) ​​​​

The statement in the question is shown in the diagram,

 tan \(\alpha={y\over x}\) and tan \(\beta ={y\over MA}={y\over R-x},\) where R is horizontal range
\(\therefore\) tan \(\alpha\)+ tan\(\beta ={y\over x}+{y\over R-x}\)
\(={(R-x+x)y\over x(R-x)}={yR\over x(R-x)}\)
or   tan \(\alpha\)+ tan\(\beta ={yR\over x(R-x)}\)       ..........(i)
Again, x =(u cos \(\theta\)) t                    ..........(ii)
y=(u sin \(\theta\)) t -\({1\over2}\)gt²                       ...........(iii)
From (ii) and (iii), we have
y = x tan  \(\theta =[1-{xg\over 2u^2 cos^2\theta tan \theta}]\)
Putting,    \(R={2u^2sin \theta cos \theta\over g}\)
we get y=\(x=tan \theta[1-{xg \over2u^2 cos\theta sin \theta}]\)
\(=xtan\theta[1-{x\over R}]\)
or \({y\over x}=tan \theta({R-x\over R})\)  .........(iv)
Putting (iv) in (i), we get
tan \(\alpha\)+ tan\(\beta ={yR\over x(R-x)} =tan \theta\)
\(\therefore\) tan \(\alpha\)+ tan\(\beta =tan \theta\).

Let the body be projected with velocity u inclined at angle \(\theta\) with the horizontal. The horizontal and vertical components of velocity and acceleration are
u_x, a_x and u_y, a_y
where u_x = u cos \(\theta\) u_y = u sin\(\theta\) ,a_x = 0,a_Y = -g
g is the acceleration due to gravity.

The coordinates of O are (0, 0) considering horizontal motion.
The position of the body after time t has coordinate (x, y);
where  x (t)=x_o + u_x t + \({1\over2}a_x t^2\)
Substituting for various factors
x (t)=x_o + U cos \(\theta\) . t + \({1\over2}\) x 0 x t²
or x (t) =u cos \(\theta\) . t
\(t={x(t)\over u \ cos \theta}\)            ............(i)
Considering the vertical motion
y (t) Y (0) + u_y +\({1\over2}a_y t^2\)
or  y (t)=0+ U sin \(\theta\) . t - \({1\over2}\) gt²
or  y (t)=U sin \(\theta\) . t - \({1\over2}\) gt²  ............(ii)
Substituting for t from equation (i) in equation (ii), we get
\(y(t)=u \ sin \theta ({x(t)\over u \cos \theta})-{1\over 2}g({x(t)\over u \ cos \theta})^2\)
\(\Rightarrow \ y(t)=x(t)tan \theta -{1\over2}g{x^2(t)\over u^2 \ cos^2 \theta}\) ...............(iii)
This is an equation of parabola. Thus, the path of a projectile is a parabola.
Maximum height attained. At the maximum height, the vertical component of velocity becomes zero. Now using the equation of motion.
\(h={v^2_y-u^2_y\over 2a_y}\)
We have maximum height
\(\therefore h_{max}={0^2-(u \ sin \theta)^2\over 2(-g)}\)
or \(H={u^2sin^2\theta \over 2g}\)  .....................(iv)
Time for maximum height. Using equation of motion v = u + at
or v_x = u_x + a_y t
we have 0 =u sin \(\theta\) - gt
or t=\(u \ sin \theta \over \ g\) .....................(v)
Horizontal range. Let the horizontal range be x. Since there is no acceleration in the horizontal direction so
x = x (0) + u_x t +  \({1\over2}a_x t^2\)
As x (0) = 0, u_x = U cos \(\theta\), a_x= 0 and it is the total time of the flight which is twice the time for maximum height because body takes same time in rising to and falling from the highest point.
Hence,   t = \({2u \ sin \theta \over g}\)
\(\therefore \) x = o + u cos \(\theta\) . t = u cos\(\theta\)\(({2u \ sin \theta \over g})\)
or x \(={u^2\over \ g}(2 \ sin \theta cos \theta)\)
\(\Rightarrow \ x={u^2\over g}sin \ 2 \theta\) ....................(vi)
Maximum horizontal range. From equation (vi) for x to be maximum, the value of sin 2\(\theta\) should be maximum which is 1,
hence         \(x_{max}={u^2\over g}\) ....................(vii)
For this x_max' sin 2\(\theta\)= 1\(\Rightarrow\) \(\theta\) = 45°
Therefore, the horizontal range will be maximum if the angle of projection is 45° or \(\pi\over 4\) radians.
Time of flight of the projectiles. The projectile after completing its flight returns back to the same horizontal level from which it was projected. Therefore, the vertical displacement in the whole flight is zero. Considering vertical motion.
y(t) = y(0)  +u_yt + \({1\over2}a_yt^2\)
Now y(t) = 0,y(o) = 0,u_y = u sin \(\theta\) , a_y = -g
Then 0 = 0 + u  sin \(\theta\) . T-\({1\over2}gT^2\)
\(\Rightarrow T({u \ sin \theta -{1\over2}gT})=0\)
Therefore T = 0
and  U sin \(\theta -{1\over 2}gT=0\Rightarrow u\ sin \theta ={1\over2}gT\)
or gT = 2u sin \(\theta\)
\(T={2u \ sin \theta \over g}\) ...........(viii)
Equation (viii) gives the total time of flight. This is twice the time for maximum height.

Given, A = 5 units, B = 5units, \(\theta =60°\) A + B=? and A - B = ?

The magnitude of the resultant vectors of the sum,
\(R=\sqrt { { A }^{ 2 }+{ B }^{ 2 }+2AB\cos { \theta } } \) 
=\(\sqrt { { 5 }^{ 2 }+{ 5 }^{ 2 }+2\times 5\times 5\times \cos { 60° } } \) 
= \(5\sqrt { 3 } \) unit
The magnitude of the resultant vector of the difference,
 \(R=\sqrt { { A }^{ 2 }+{ (-B) }^{ 2 }+2AB\cos { \theta } } \)
R =\(\sqrt { { 5 }^{ 2 }+{ 5 }^{ 2 }+2\times 5\times 5\times \cos { 120° } } \) 
R = 5 unit

Given, \(\nu _{ 0 }=5\overset { \wedge }{ i } m/s\),a = \(\nu _{ 0 }=3\overset { \wedge }{ i } +2\overset { \wedge }{ j } m/s^{ 2 }\)
y(t) = ?,x(t) = 84m (ii) Speed \(\nu \) = ?
Then, \(y(t)={ \nu }_{ 0 }\quad t+\frac { 1 }{ 2 } at^{ 2 }\quad and\quad { \sqrt { { v }_{ x }^{ 2 }+{ v }_{ y }^{ 2 } } }\)
\(y(t)=5\overset { \wedge }{ i } t+\frac { 1 }{ 2 } (3\overset { \wedge }{ i } +2\overset { \wedge }{ j } ){ t }^{ 2 }\)
\(\\ =\left( { 5\overset { \wedge }{ i } }t+\frac { 3 }{ 2 } { t }^{ 2 } \right) \overset { \wedge }{ i } +{ t }^{ 2 }\overset { \wedge }{ j } \)
On comparing, \(\quad x(t)=5t+\frac { 3 }{ 2 } { t }^{ 2 }\)
\(\\ \Rightarrow y(t)=1{ t }^{ 2 }\) 

The path followed by a projectile projected at an angle \(\theta\) with velocity \(\overset\rightarrow{u}\) is shown in figure. The maximum height attained by the projectile is given by
H=\(\frac { { u }^{ 2 }{ sin }^{ 2 }{ \theta }_{ 0 } }{ 2g } \)   ..(i)
The range of the projectile is given by
R=\(\frac { { u }^{ 2 }{ sin }^{ 2 }{ \theta }_{ 0 } }{ g } =\frac { { 2u }^{ 2 }{ sin }^{ 2 }{ \theta }_{ 0 }cos{ \theta }_{ 0 } }{ g } \)      ..(ii)
Dividing eqn. (i) by eqn. (ii), we get
tan \({ \theta }_{ 0 }=\frac { 4H }{ R } \)
\(\Rightarrow\) \(\theta\)₀ tan^-1\(\left( \frac { 4H }{ R } \right) \) .

Consider a projectile projected at an \(\theta\) angle  to the horizontal with velocity u, the horizontal and vertical components initially with velocity u cos \(\theta\) and u sin \(\theta\) respectively. Vertical velocity at highest point is zero, due to gravity acting vertically downwards.
Using, \(\upsilon \)=u+at
we have, 0=u sin \(\theta\) -gt
\(\Rightarrow\)  t=\(\frac { u sin\theta }{ g } \)
The time to reach topmost point, t =\(\frac { u sin\theta }{ g } \)
Using \(\upsilon ^{ 2 }\)=u²+2as
we have, 0=u²sin²\(\theta\)-2g h_max
\(\Rightarrow\) h_max=\(\frac { { u }^{ 2 }{ sin }^{ 2 }\theta }{ 2g } \).

(i) False, each component of a vector is also a vector.
(ii) True, because the total path length is either greater than or equal to the magnitude of the displacement vector.

For a projectile launched with velocity \(v _{ 0 }\)at an angle \(\theta _{ 0 }\), the range is given by \(R=\frac { { v }_{ 0 }^{ 2 }sin2\theta _{ 0 } }{ g } \).
Now, for angles, (45° + α) and ( 45° – α), 2θ_o is (90° + 2α) and ( 90° – 2α) , respectively. The values of sin (90° + 2α) and sin (90° – 2α) are the same, equal to that of cos 2α. Therefore, ranges are equal for elevations which exceed or fall short of 45° by equal amounts α.

Unit vector perpendicular to both
\(\overrightarrow{\mathrm{A}}=2 \hat{i}+\hat{j}+\hat{k} \text { and } \hat{i}-\hat{j}+2 \hat{k}\)
is given by \(\hat{n}=\frac{\overrightarrow{\mathrm{A}} \times \overrightarrow{\mathrm{B}}}{|\overrightarrow{\mathrm{A}} \times \overrightarrow{\mathrm{B}}|}\)
\(=\overrightarrow{\mathrm{A}} \times \overrightarrow{\mathrm{B}}=\left|\begin{array}{rrr} \hat{i} & \hat{j} & \hat{k} \\ 2 & 1 & 1 \\ 1 & -1 & 2 \end{array}\right|\)
\(=\hat{i}[2-(-1)]-\hat{j}(4-1)+\hat{k}(-2-1)\)
\(=3 \hat{i}-3 \hat{j}-3 \hat{k}\)
Unit vector is \(\hat{n}=\frac{3 \hat{i}-3 \hat{j}-3 \hat{k}}{\sqrt{9+9+9}}\)
  \(\frac { 3\hat { i } -\hat { 3j } +3\hat { k } }{ \sqrt { 27 } } \)