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

It is a uniform circular motion. So the direction of velocity changes but not the magnitude. Therefore speed in considered constant. Again magnitude of acceleration does not change.

For freely falling body, velocity while the body, hit the ground \(v=\sqrt{2 g h}\)
\(v_{1}=\sqrt{2 g h_{1}} \text { and } v_{2}=\sqrt{2 g h_{2}} \)
\(\therefore \frac{m_{1} v_{1}}{m_{2} v_{2}}=\frac{m_{1} \sqrt{h_{1}}}{m_{2} \sqrt{h_{2}}}=\frac{m_{1}}{m_{2}} \sqrt{\frac{h_{1}}{h_{2}}}\)

Total distance to be covered = 1 km + 100 m = 1100 m (including both bridge and time)
Then, Speed=60 kmh^-1\(=60\times\frac{5}{18}ms^{-1}=\frac{50}{3}\ ms^{-1}\)
Then, time taken to cover this distance \(=\frac{1100}{\frac{150}{9}}s=66s\)

The total distance the athlete covered = 3\(\times\)circumference of track
Distance = 3\(\times\) 2\(\pi\)\(\times\) 50 = 300\(\pi\) m (or)
Distance = 300\(\times\)3.14 = 942 m
The displacement is zero, since the athlete reaches the same point A after three rounds from where he started.

\(\vec { \tau } =\vec { r } \times \vec { F } =\left| \begin{matrix} \hat { i } & \hat { j } & \hat { k } \\ 2 & 3 & 5 \\ 3 & -2 & 4 \end{matrix} \right| \)
\(\vec { \tau } =\left( 12-\left( 10 \right) \right) \hat { i } +\left( 15-8 \right) \hat { j } +\left( -4-9 \right) \hat { k } \)
\(\vec { \tau } =22\hat { i } +7\hat { j } -13\hat { k } \)

 \(\vec { { r }_{ 1 } } =3\hat { i } +4\hat { j } \)
\(\vec { { r }_{ 2 } } =\hat { i } +2\hat { j } \)    
\(\Delta \overrightarrow { r } =\overrightarrow { { r }_{ 2 } } -\overrightarrow { { r }_{ 1 } } \)

\(=(\hat { i } +2\hat { j } )-(3\hat { i } +4\hat { j } )\)
\(\Delta \overrightarrow { r } =-2\hat { i } +\hat { j } \)

The acceleration of a particle at any instant 'r' is equal to the rate of change of velocity.
Acceleration \(\overrightarrow{a}=\frac{d\overrightarrow{v}}{dt}\). It is a vector quantity.

The linear speed at t = 2 s
v = 3t = 6 ms^-1
The centripetal acceleration at t = 2 s is
\(a_c=\frac{v^2}{r}=\frac{(6)^2}{10}=3.6\ ms^{-2}\)
The tangential acceleration is \(a_t=\frac{dv}{dt}=3\ ms^{-2}\)
The angle between the radius vector with resultant acceleration is given by
\(tan\theta=\frac{a_t}{a_c}=\frac{3}{3.6}\)= 0.833
\(\theta=tan^{-1}(0.833)\) = 0.69 radian
In terms of degree \(\theta=0.69\times57.17^0\approx40^0\)

The motion of the cricket ball in air is essentially a projectile motion. As we have already seen, the range (horizontal distance) of the projectile motion is given by
\(R=\frac{u^2sin2\theta}{g}\)
The initial speed u = 30 ms^-1
The projection angle θ = 30°
The horizontal distance travelled by the cricket ball  \(R=\frac{(30)^2\times sin60^0}{10}=\frac{900\times\frac{\sqrt 3}{2}}{10}=77.94\ m\)
This distance is greater than the distance of the boundary line. Hence the ball will cross this line and go for a six.

We know that speed is the magnitude of the velocity vector. Hence,
Speed of A = \(|\vec{V_A}|=\sqrt{(3)^2+(-5)^2+(2)^2}=\sqrt{9+25+4}=\sqrt{48}ms^{-1}\)
Speed of B = \(\vec{V_B}=\sqrt{(1)^2+(2)^2+(3)^2}=\sqrt{1+4+9}=\sqrt{14}ms^{-1}\)
Spedd of C = \(\vec{V_C}=\sqrt{(5)^2+(3)^2+(4)^2}=\sqrt{25+9+16}=\sqrt{50}ms^{-1}\)
The particle C has the greatest speed.
\(\sqrt{50}>\sqrt{38}>\sqrt{14}\)

(a) v_x - remains constant
(b) v_y - decreases and increases
(c) a - remains downward
(d) r - varies