Range is same for the angle if projection \(\theta \text { and } 90-\theta\)

\(\text {Time of flight }=\frac{2 u}{g}\)

For a free falling body, initial vertical downward velocity = 0
For a body thrown horizontally with a velocity, initial vertical downward velocity = 0
Since both having same initial vertical downward velocity they cover equal vertical distance at any instant

\(\vec{v}=2 \hat{l}+t^{2} \hat{j}-9 \vec{k}\)
\(\vec{a}=\frac{d \vec{v}}{d t}=2 t \hat{j}\)
\(\text { When } t=0.5 \mathrm{~s}\)
\(a=1 \mathrm{~ms}^{-2}\)

Answers (a), (b) and (c) are all anticlockwiseand answer (d) alone is in the clockwise direction

In projectile motion, the range of particle is given by,
\(R=\frac{u^2sin\ 2\theta}{g}\)
\(\theta=\frac{\pi}{4}\ u=v_0=10\ ms^{-1}\)
\(\therefore\ R_{earth}=\frac{(10)^2\sin\frac{\pi}{2}}{9.8}=\frac{100}{9.8}\)
R_earth = 10.20 m (Approximately 10 m)
If the same object is thrown in the Moon, the range will increase because in the Moon, the acceleration due to gravity is smaller than g on Earth,
\(g_{moon}=\frac{g}{6}\)
\(R_{moon}=\frac{u^2sin2\theta}{g_{moon}}=\frac{v_0^2sin2\theta}{\frac{g}{6}}\)
\(\therefore\ R_{moon}=6\times10.24=61.22m\) (Approximately 60 m)
The range attained on the Moon is approximately six times that on Earth.

1. \(\omega ={ \omega }_{ 0 }+\alpha t\)	\(\omega \) = Final angular velocity
2. \(\theta ={ \omega }_{ 0 }t+\frac { 1 }{ 2 } { \alpha t }^{ 2 }\)	\({ \omega }_{ 0 }\) = initial angular velocity
3. \({ \omega }^{ 2 }={ \omega }_{ 0 }^{ 2 }+2\alpha \theta \)	\(\theta \) = Angular displacement
4. \(\theta =\frac { \left( { \omega }_{ 0 }+\omega \right) t }{ 2 } \)	\(\alpha \) = angular acceleration t = time

Angular displacement: While a particle is revolving around a point in a circular path, the angle described by the particle about the axis of rotation (or at the centre of the circle) in a given time is called angular displacement. Its unit is radian.
Angular velocity: The rate of change of angular displacement is called angular velocity. Its unit is rad s^-1
\(\omega =\frac { d\theta }{ dt } \)

Velocity:
Velocity is equal to the rate of change of position vector with respect to time. 
It is a vector quantity \(\overrightarrow{v}=\frac{d\overrightarrow{r}}{dt}\)
Speed:
The magnitude of velocity is called speed and is given by \(v= \sqrt{v^2_x+v^2_y+v^2_z}\). It is a positive scalar.

\(u =5 m s^{-1}  \)
\(\theta =30^{\circ}  \)
\(h_{\max } =\frac{u^{2} \sin ^{2} \theta}{2 g}=\frac{25 \times\left(\frac{1}{2}\right)^{2}}{2 \times 9.8}=0.318 \mathrm{~m}  \)
\(\text { Range } =\frac{u^{2} \sin 2 \theta}{g}=\frac{25 \times \sqrt{3}}{9.8 \times 2}=2.21 \mathrm{~m}\)

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

When a point object is moving on a circular path if the velocity of the object changes both in speed (magnitude) and direction the motion is said to be non uniform circular motion.

Consider a non uniform circular motion of a point object moving on a circular path of radius r with a velocity v. In a non uniform circular motion the magnitude of the velocity (speed) is not constant. Whenever the speed is not same in circular motion, the object will have both centripetal and tangential acceleration as shown in the figure.
The resultant acceleration is obtained by vector sum of centripetal and tangential acceleration. Since the centripetal acceleration is \(\frac{v^2}{r}\) the magnitude of the resultant acceleration is given by \(a_R =\sqrt{a^2_t+(\frac{v^2}{r})}^2\)
The resultant acceleration makes an angle 0 with the radius vector as shown in the figure. The angle is given by \(tan \ \theta=\frac{a_t}{({v^2}/{r})}\)

Consider a uniform circular motion of a point object moving on a circular path of radius r. In uniform circular motion, the velocity vector turns continuously without changing its magnitude (speed). Let v be its velocity.
The velocity is tangential at every point in the circle. But the acceleration 'a' is acting towards the centre of the circle. This is called centripetal acceleration. It always points towards the centre of the circle.
Let the direction of position and velocity vectors shift through the same angle θ in a small interval of time \(\triangle t\) as shown in figure. For uniform circular motion \(r=|\overrightarrow{r_1}|={r_2}| \) and \(vr=|\overrightarrow{v_1}|=\overrightarrow{v_2}|\)

If the particle moves from position vector \(\overrightarrow{r_1}\) to \(\overrightarrow{r_2}\) , the displacement is \(\triangle{\overrightarrow{r}}={\overrightarrow{r}}_{2}-{\overrightarrow{r}}_{1}\) and change in velocity from \(\overrightarrow{v_1}\) to \(\overrightarrow{v_2}\) is \(\triangle={\overrightarrow{v}}_{1}={\overrightarrow{v}}_{2}-{\overrightarrow{v}}_{1}\) then the magnitude of the displacement \(\triangle r \ and \ of \triangle v\) satisfy the following relation:
\({{\triangle r}\over{r}}=-{{\triangle v}\over{v}}=\theta\)
Here the negative sign implies that \(\triangle r\) points radially inward, towards the centre of the circle
\(\triangle v=-v.{{\triangle r}\over{r}}\)
Then \(a={{\triangle r}\over{\triangle t}}={{r}\over{r}}\left( {{\triangle r}\over{\triangle t}} \right)=-{{v}^{2}{}\over{r}}\)
For uniform circular motion v = \(\omega\)r. where \(\omega\) - angular velocity of the object about the centre. Then the centripetal acceleration can be written as
\({a=-\omega^2 r}\)

Consider an object thrown with initial velocity \(\overrightarrow{u}\) at an angle \(\theta\) with the horizontal.
Then,
\(\overrightarrow{u}=u,\overrightarrow{i}+u,\overrightarrow{j}\)
where u_x =u cos \(\theta\) is the horizontal component and u_y = u sin \(\theta\) the vertical component of velocity.

Since the acceleration due to gravity acts in the direction opposite to the direction of vertical component u_y, this component will gradually reduce to zero at the maximum height of the projectile. At this maximum height, the same gravitational force will push the projectile to move downward and fall to the ground.
But, there is no acceleration along the x direction throughout the motion. So, the horizontal component of velocity (u_x = u cos \(\theta\)) remains the same till the object reaches the ground.
After anytime t, the velocity along horizontal motion
v_x = u_x + a_xt = u_x = u cos\(\theta\)  [∴ a_x = 0]
The horizontal distance travelled by projectile in time t is
s_x = \({u}_{x}t+{{1}\over{2}}a_n{t}^{2}\)
[∴ s_x = x and a_n = 0]
∴ t = \(\frac{x}{u \cos \theta} \) ......(1)
For the vertical motion the velocity after time t is v_y = u_y + a_yt
v_y = u sin \(\theta\) - gt........(2)
(∴ u_g = u sin \(\theta\) and a_y = - g)
The vertical distance travelled by the projectile in the same time r is
s_y = \({u}_{y}t{{1}\over{2}}{a}_{y}{t}^{2}\)
y = sin \(\theta\ t-{{1}\over{2}}{gt}^{2}\)
(Here, (Here, s_y = y, u_y = u sin \(\theta\) and a_y = - g) 
Substitute the value of t from equation (1) in equation (3), we have
\(y=u\ \sin\theta{{x}\over{u\ \cos\ \theta}}-{{1}\over{2}}g{{x^2}\over{u^2{cos}^{2}\theta}}\)
\(y=x\ \tan\theta-{{1}\over{2}}g{{x^2}\over{u^2{cos}^{1}\theta}}\)
Thus the path followed by the projectile is an inverted parabola.
2. Maximum height: The maximum vertical distance travelled by the projectile during its journey is called maximum height. 
For the vertical part of the motion. \({v}_{y}^{2}={u}_{y}^{2}+2{a}_{y}s\)
Here, u_y = u sin \(\theta\),  a_y = -g, S = h_max and at the maximum height v_y = 0.
Hence,
∴ 0 = u² sin² = 2gh_max
\({h}_{max}={{{u}^{2}{sin}^{2}\theta}\over{2g}}\)
3. Horizontal Range (R): The maximum horizontal distance between the point of projection and the point on the horizontal plane where the projectile hits the ground is called horizontal range (R). 
Range R = Horizontal component of velocity x time of flight
= u cos \(\theta\times{T}_{f}\)
\(R=u\ \cos\theta\times{{2u\ \sin\theta}\over{g}}={{2{u}^{2}\sin\theta\cos\theta}\over{}g}\)      \([\therefore T_f=\frac {2u\ sin \theta}{g}]\)
\(\therefore\ R={{u^2\sin 2\theta}\over{g}}\)