(a) If they act at opposite direction, resultant is zero.
(b) If they act in same direction, R = 2F.
(c) For the resultant to be F,
F ² = F ² + F ² + 2F² cos θ
cosθ = -1/2 or θ = 120°.

In figure, 0 is the observation point at the ground, A and B are the positions of aircraft for which \(\angle AOB=30°\). Draw a perpendicular OC on AB.Here OC = 3400 m and \(\angle AOC=\angle COB=15°\).Time taken by aircraft from A to B is 10 s.
In \(\Delta \)AOC, AC=OC \(\tan { 15° } \)
= 3400\(\times \)0.2679  
= 910.86 m
AB = AC + CB = AC + AC = 2AC
=2\(\times \)910.86 m
Speed of the aircraft 
v = \(\frac { distance \ AB }{ time } =\frac { 2\times 910.86 }{ 10 } \)
= 182.17 ms^-1 = 182.2 ms^-1

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,  \(\frac { a }{ g } =\frac { { (250) }^{ 2 } }{ 1000 } \times\frac { 1 }{ 9.8 } =6.38\)

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.

\(x=0+\frac{1}{2} \times 9.8 t^{2}=4.9 t^{2}\)                 .............(i)
\(100-x=25 t+\frac{1}{2} \times(-9.8) t^{2}\)            ...............(ii)

Solving equation (i) and (ii), we get
t = 4 second
x = 4.9 x 16 = 78.4 m

We know that s = r θ if a body covers an arc of length s in a radius r, turning its radial line by θ. Differentiating both sides with respect to time, we have
\(\frac{d s}{d t}=r \frac{d \theta}{d t} \quad \text { i.e., } \quad v=r \omega .\)
Linear velocity = radius x angular velocity.
At each point the body moves along the tangent. The presence of centripetal force mv²/r makes it to pass in the circular path. Thus the direction of velocity is always along the tangent at any point in the circular path.

\(\mathbf{v}(t)  =\frac{\mathrm{d} \mathbf{r}}{\mathrm{d} t}=\frac{\mathrm{d}}{\mathrm{d} t}\left(3.0 t \hat{\mathbf{i}}+2.0 t^2 \hat{\mathbf{j}}+5.0 \hat{\mathbf{k}}\right) \)
\(=3.0 \hat{\mathbf{i}}+4.0 t \hat{\mathbf{j}} \)
\(\mathbf{a}(t)  =\frac{\mathrm{d} \mathbf{v}}{\mathrm{d} t}=+4.0 \hat{\mathbf{j}} \)
\(a  =4.0 \mathrm{~m} \mathrm{~s}^{-2} \text { along } y \text {-direction }\)
At \(t=1.0 \mathrm{~s}, \quad \mathbf{v}=3.0 \hat{\mathbf{i}}+4.0 \hat{\mathbf{j}}\)
It's magnitude is \(v=\sqrt{3^2+4^2}=5.0 \mathrm{~m} \mathrm{~s}^{-1}\) and direction is \(\theta=\tan ^{-1}\left(\frac{v_y}{v_x}\right)=\tan ^{-1}\left(\frac{4}{3}\right) \equiv 53^{\circ} \text { with } x \text {-axis. }\)

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=\(\frac { { u }^{ 2 }{ sin }^{ 2 }\theta }{ 2g } \).

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
\(H=\frac { { u }^{ 2 }{ sin }^{ 2 }\theta }{ 2g } \Rightarrow 25=\frac { { (40) }^{ 2 }{ sin }^{ 2 }\theta }{ 2\times 9.8 }\)
\(or\ { \sin }^{ 2 }\theta =\frac { 25\times 2\times 9.8 }{ 1600 } =0.3068\)
\(or\ \sin\theta =0.5534=\sin{ 33.6 }^{ 0 }\)
\(or\ \theta ={ 33.6 }^{ 0 }\)
\( \therefore \text{ Horizontal range (R)}=\frac { { u }^{ 2 }\sin2\theta }{ g }\)
\( \\ =\frac { { (40) }^{ 2 }sin2\times { 33.6 }^{ 0 } }{ 9.8 } =\frac { 1600\times sin{ 67.2 }^{ 0 } }{ 9.8 } \)
\(\\ =\frac { 1600\times 0.9219 }{ 9.8 } =150.5m\)

(a) True, The magnitude of a vector is a number. Hence, it is a scalar.
(b) False, each component of a vector is always a vector, not scalar.
(c) False ,Total path length is a scalar quantity, whereas displacement is a vector quantity. Hence, the total path length is always greater than the magnitude of displacement. It becomes equal to the magnitude of displacement only when a particle is moving in a straight line.
(d) True, because the total path length is either greater than or equal to the magnitude of the displacement vector.
(e) True, this is because the resultant of two vectors will not lie in the plane of third vector and hence cannot cancel its effect to give null vector.

(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.

(i) Projectile motion is independent of mass, it only depends on angle of projection and initial velocity of projection.
(ii) The maximum height attained by projectile is equal to one fourth of its maximum range.
(iii) At the highest point of projectile path the component of velocity becomes zero, therefore angle between acceleration acting vertically downward and horizontal component of velocity becomes 90°.
(iv) Since projectile follow a prabolic path, it strike back on same horizontal plane with same initial velocity and with same angle at which it was projected.
(v) At complimentary angles of projection i.e., θ and 90 - θ, the horizontal range is same.
(vi) \(y=x \tan \theta-\frac{g x^{2}}{2 u^{2} \cos ^{2} \theta}\)