Yes, adding a component of a vector of the same vector is meaningful because both vectors are of same dimensions

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

No adding a scalar to a vector of the same dimensions is not meaningful because a scalar cannot be added to a vector

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 \(u=[2g(b+{a^2\over 16b})]^{1/2}\) ​​​​

For the definition of angular velocity and angular acceleration, see text.
Numerical. Velocity at highest point = u cos \(\theta\) = V₁ (given)
\(h_{max}={u^2sin^2\theta\over 2g}\)
Vertical velocity at \({h_{max}\over 2}=V_{2y}=\sqrt{u^2sin^2\theta 2g{h_{max}\over 2}}\)
or \(V_{2y}=\sqrt{u^2sin^2\theta (1-{1\over 2})}\)
\(={u \ sin \theta\over \sqrt{ 2}}\)
V_2x  = u cos \(\theta\)
\(V_2=\sqrt{V^2_{2x}+V^2_{2y}}\)
\(=\sqrt{u^2cos^2\theta +{u^2 \ sin^2 \theta\over { 2}}}\)
Given, \(V_1=\sqrt{6\over 7}V_2\)
\(\therefore {u \ cos \theta \over u\sqrt{cos^2\theta+{sin^2 \theta\over2}}}=\sqrt{6\over 7}\)
Squaring both the sides
\( { \ cos^2 \theta \over {cos^2\theta+{sin^2 \theta\over2}}}={6\over 7}\)
or \( {1 \over {1+{tan^2 \theta\over2}}}={6\over 7}\)
or \(1+{tan^2 \theta\over2}={7\over 6}\Rightarrow{tan^2 \theta\over2}={7\over 6}-1={1\over6}\)
or \(tan \theta =\sqrt{2\over 6}={1\over \sqrt{3}}\)
\(\therefore \ \theta =tan^{-1}({1\over \sqrt{3}})=30^o\).

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\(\beta =tan \theta\).

Consider motion of a particle in x-y plane.
Displacement. Let the particle be at position A having position vector \(\vec{r_1}\) at time t₁ where \(\vec{r_1}=(x_1\hat{i}+y_1\hat{j}).\)  At time t₂, the object reaches point B having position vector\(\vec{r_2}\) where \(\vec{r_2}=(x_2\hat{i}+y_2\hat{j}).\)
\(\therefore\) Displacement of the particle during the time (t₂ - t₁) will be
\(\vec{r}\) or \(\vec{r_{12}}=\vec{r_2}-\vec{r_1}=(x_2-x_1)\hat i +(y_2-y_1)\hat{j}, \ where | \vec{r}|=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}\)
Velocity. If motion be uniform then uniform velocity may be defined as the displacement covered per unit time i.e.,
Velocity \(\vec{v}={\triangle \overrightarrow {r} \over t }={\overrightarrow{r_2}-\overrightarrow {r_1}\over t_2 -t_1}={(x_2-x_1)\over(t_2-t_1)}\hat{i}+{(y_2-y_1)\over(t_2-t_1)}\hat{j}=v_x\hat{i}+v_y\hat{j}\)
where V_x=\({x_2-x_1\over t_2-t_1},v_y={y_2-y_1\over t_2-t_1}and |\vec{v}|=\sqrt{v^2_x+v^2_y}\) .
Acceleration. If velocity of particle is variable then time rate of change of velocity is the acceleration \(\vec{a}\) . If \(\vec{v_1}\) and \(\vec{v_2}\) be the velocities of the particle at times t₁ and t₂ respectively and acceleration be uniform, then

\(\vec{a}={\triangle \vec{v}\over \triangle t}={\vec{v_2}-\vec{v_1}\over t_2-t_1}\)
\(={(v_{2x}\hat{i}+v_{2y}\hat{j})-(v_{1x}\hat{i}+v_{1y}\hat{j})\over (t_2-t_1)}\)
\(={(v_{2x}-v_{1x})\over (t_2-t_1)}\hat{i}+{(v_{2y}-v_{1y})\over (t_2-t_1)}\hat{j}\)
\(=a_x\hat{i}+a_y\hat{j}\)
where \(a_x={v_{2x}-v_{1x}\over t_2-t_1},a_y={v_{2y}-v_{1y}\over t_2-t_1} \ and \ |\vec{a}|=\sqrt{a^2_x+a^2_y}\)

(i)  \(\hat { i } \) + \(\hat { j } \) =\(\sqrt { \left( 1 \right) ^{ 2 }+\left( 1 \right) ^{ 2 }+2\times 1\times 1\times cos\quad 90^{ 0 } } =\sqrt { 2 } \) = 1.414 units
tan\(\theta \)  = \(\frac{1}{1}\)=1,  \(\therefore\) \(\theta \) = 45°
So the vector  \(\hat { i } \) + \(\hat { j } \) makes an angle of 45° with x-axis.
(ii) \(\left| \hat { i } -\hat { j } \right| \sqrt { \left( 1 \right) ^{ 2 }+\left( 1 \right) ^{ 2 }-2\times 1\times 1\times cos\quad 90^{ 0 } } \)
=\(\sqrt{2}\) = 1.414 units
The vector  \(\hat { i } \) - \(\hat { j } \)makes an angle of -45° with .r-axis.
(iii) Let us now determine the component of \(\overset { \rightarrow }{ A } \)= 2\(\hat { i } \)+3\(\hat { j } \)in the direction of  \(\hat { i } \) + \(\hat { j } \) .
Let \(\overset { \rightarrow }{ B } \) = \(\hat { i } \) + \(\hat { j } \)
\(\overset { \rightarrow }{ A } \).\(\overset { \rightarrow }{ B } \) cos \(\theta \) =(A cos \(\theta \))B
So the component of \(\overset { \rightarrow }{ A } \) in the direction of \(\overset { \rightarrow }{ B } \) =\(\frac { \overset { \rightarrow }{ A } .\overset { \rightarrow }{ B } }{ B } \)
=\(\frac { \left( 2\hat { i } +3\hat { j } \right) .\left( \hat { i } +\hat { j } \right) }{ \sqrt { \left( 1 \right) ^{ 2 }+\left( 1 \right) ^{ 2 } } } =\frac { 2\hat { i } .\hat { i } +2\hat { i } .\hat { j } .\hat { i } +3\hat { j } .\hat { j } }{ \sqrt { 2 } } =\frac { 5 }{ \sqrt { 2 } } \)units
( iv) Component of \(\overset { \rightarrow }{ A } \) in the direction of  \(\hat { i } \) - \(\hat { j } \)=\(\hat { i } -\hat { j } =\frac { \left( 2\hat { i } +3\hat { j } \right) .\left( \hat { i } -\hat { j } \right) }{ \sqrt { 2 } } =-\frac { 1 }{ \sqrt { 2 } } \)units.

Let the ball strike the nth step of stairs.
∴ Vertical distance travelled \(=\mathrm{ny}=\mathrm{n} \times 0.2=\frac{1}{2} \mathrm{gt}^2\)
Horizontal distance travelled \(=\mathrm{nx}=\mathrm{ut}\)
\(\Rightarrow \mathrm{t}  =\frac{\mathrm{nx}}{\mathrm{u}} \)
\(\mathrm{ny}  =\frac{1}{2} \mathrm{gt}^2=\frac{1}{2} \mathrm{~g}\left(\frac{\mathrm{nx}}{\mathrm{u}}\right)^2=\frac{1}{2} \mathrm{~g} \frac{\mathrm{n}^2 \mathrm{x}^2}{\mathrm{u}^2} \)
\(\Rightarrow \mathrm{n}  =\frac{2 \mathrm{u}^2 \mathrm{y}}{\mathrm{gx^{2 }}} \)
\(=\frac{2 \times 1.8^2 \times 0.2}{9.8 \times 0.2^2} \approx 3.3\)
So, Fourth step.

\(\sqrt { { A }^{ 2 }+B^{ 2 }+AB } \)