Motion in Two Dimensions

Define velocity vector.

A volleyball is served up and to the right at an angle $30°$with a speed of $24.3 \mathrm{m}/\mathrm{s}$. Determine the vertical and horizontal components of the velocity at the moment shown.

Define velocity vector.

Define position vector.

What is the difference between in dimensional and two-dimensional motion?

Suppose an object is at point $A$ at time $=$ $0$ and at point $B$ at time $= t$. The position vectors of the object at point $A$ and point $B$ are given as

$$\begin{gathered} \widehat{r_A}=6 \hat{i}+5 \hat{j}+3 \hat{k} \\ \widehat{r_B}=3 \hat{i}+4 \hat{j}+2 \hat{k} \end{gathered}$$

Find the displacement vector of the particle from the time interval $0$ to $t$.

Suppose an object is at point $A$ at time $=$ $0$ and at point $B$ at time $= t$. The position vectors of the object at point $A$ and point $B$ are given
as

$$\begin{gathered} \widehat{r_A}=5 \hat{i}+3 \hat{j}+4 \hat{k} \\ \widehat{r_B}=2 \hat{i}+2 \hat{j}+1 \hat{k} \end{gathered}$$

Find the displacement vector of the particle from the time interval $0$ to $t$.

An enemy plane is flying horizontally at an altitude of $2 km$ with a speed of $300 m{s}^{-1}$. A soldier with an anti-aircraft tanker on the ground sights the enemy plane when it is directly overhead and fires a shell with a muzzle speed of $600 m{s}^{-1}$. At what angle with the vertical should the shell be fired so as to hit the plane? 

An arrow is shot from ground with velocity $20\sqrt{2}m{s}^{-1}$ at an angle $45°$. At what time arrow is at height $15 m$ from ground? Take $(g=10m{s}^{-2})$

 A particle from its starting point if it undergoes two successive displacement given by $\overrightarrow{s_1}=20 m, 37° North of West, \overrightarrow{s_2}=50 m, 53°North of East.$Then the net displacement would be

Velocity of the particle changes from $\overrightarrow{v_1}=2\hat{i}+3\hat{j} (m/s) \mathrm{to} \overrightarrow{{\mathrm{v}}_2}=10\hat{\mathrm{i}}+9\hat{\mathrm{j}} (\mathrm{m}/\mathrm{s}) \mathrm{in} 5 \mathrm{s}$. Then the magnitude of change in velocity would be

A particle is dropped from the same height when another particle is thrown horizontally. Both the particles will hit the ground: 

Two velocities, each $5 {\mathrm{ms}}^{-1}$ are inclined to each other at $60°$. Find their resultant. Write the answer in terms of degrees.

Find the value of n for which vector $\left(\hat{i}+5\hat{j}+n\hat{K}\right)is$ perpendicular to vector $\left(2\hat{i}-\hat{j}+\hat{K}\right)$

A person moves $30\mathrm{m}$ north, then $20\mathrm{m}$ east and finally $30\sqrt{2}\mathrm{m}$ south west. What is the displacement from original position ?

If vectors $\overrightarrow{A}=\hat{\mathrm{i}}+3\hat{\mathrm{j}}+2\hat{\mathrm{k}}$ and $\overrightarrow{B}=3\hat{\mathrm{i}}+\hat{\mathrm{j}}+2\hat{\mathrm{k}}$ then find value of $\overrightarrow{A}\times \overrightarrow{B}$..

Convert the vector $\overrightarrow{r}=3\stackrel{\wedge }{i}+2\stackrel{\wedge }{j}$ into a unit vector.

The position of a particle is given by $\overrightarrow{r}=\left(3t\hat{i}+4t^3\hat{j}+5\hat{k}\right)\mathrm{m}$. Find the magnitude of velocity and acceleration of particle at $t=5s$.

For the given values of acceleration vector, what will be the resultant magnitude of the acceleration?

$a_x= 11 m/s^2, a_y= 13 m/s^2$.

Give the answer in terms of $\mathrm{m}/{\mathrm{s}}^2$.

The vector joining the points $A\left(1, 1,-1\right)$ and $B\left(2,-3, 4\right)$ and pointing from $A$ to $B$ is

If  $\left(\stackrel{\rightharpoonup }{V_1}\right) = 3\widehat{i } +4\widehat{j }+\hat{k} and \left(\stackrel{\rightharpoonup }{V_2}\right) = \widehat{i } -\widehat{j }-\hat{k}$then determine magnitude of $\left(\stackrel{\rightharpoonup }{V_1}\right) +\left(\stackrel{\rightharpoonup }{V_2}\right)$.