Motion in a Plane

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New answer posted•a year ago

Motion in a Plane Class 11th

A projectile is projected with velocity of 25m/s at an angle θ with the horizontal. After t seconds its inclination with horizontal becomes zero. If R represents horizontal range of the projectile, the value of θ will be :

[Use g = 10 m/s²]

8 Views|Posted a year ago

Asked by Vishal Baghel

Answered by

Vishal Baghel | Contributor-Level 10

a year ago

T = 2t = $\frac{}{} \frac{}{}$ $\frac{2 u s i n θ}{g} \Rightarrow t = \frac{u s i n θ}{g}$

R = u cos θ (T)

$\Rightarrow R = u c o s θ (\frac{2 u s i n θ}{g})$

$\Rightarrow R = u c o s θ 2 t$

$\Rightarrow c o s θ = \frac{R}{50 t} \times \frac{t}{t}$

$\Rightarrow c o s θ = \frac{R t}{50 t^{2}}$

$\Rightarrow c o s θ = \frac{R}{50 t^{2}} \times \frac{u s i n θ}{g}$

$\Rightarrow c o t θ = \frac{R}{50 t^{2}} \times \frac{25}{10}$

cot θ = $\frac{R}{20 t^{2}}$ $θ = c o t^{- 1} (\frac{R}{20 t^{2}})$ $\frac{}{^{}}$

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New answer posted•a year ago

Motion in a Plane Physics NCERT Exemplar Solutions Class 12th Chapter Four

Time period of a simple pendulum in a stationary lift is ‘T’. If the lift accelerated with $\frac{g}{6}$ vertically upwards then the time period will be:

(Where g = acceleration due to gravity)

5 Views|Posted a year ago

Asked by Payal Gupta

Answered by

Payal Gupta | Contributor-Level 10

a year ago

$g_{e f f} = g + a$

$= g + \frac{g}{6}$

$= \frac{7 g}{6}$

$T' = 2 π \sqrt[]{\frac{l}{g_{e f f}}}$

$T' = \sqrt[]{\frac{6}{7}} T$

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New answer posted•a year ago

Motion in a Plane Physics NCERT Exemplar Solutions Class 12th Chapter Four

Two vectors $\vec{A} a n d \vec{B}$  have equal magnitudes. If magnitude of  $\vec{A} + \vec{B}$ is equal to two times the magnitude of  $\vec{A} - \vec{B},$ then the angle between $\vec{A} a n d \vec{B}$  will be:

8 Views|Posted a year ago

Asked by Alok Kumar Singh

Answered by

Alok Kumar Singh | Contributor-Level 10

a year ago

$| \vec{A} + \vec{B} | = 2 | \vec{A} - \vec{B} |$

Given A = B

Squaring equation (1) both side.

${| \vec{A} + \vec{B} |}^{2} = {(2 | \vec{A} - \vec{B} |)}^{2}$

$A^{2} + B^{2} + 2 \vec{A} . \vec{B} = 4 (A^{2} + B^{2} - 2 \vec{A} . \vec{B})$

2A² + 2A² cosq = 4 (2A² – 2A² cosq)

2A² + 2A² cosq = 8A² – 8A² cosq

10A² cosq = 6A²

cosq = $\frac{3}{5}$

q = cos^-1 (3/5)

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New answer posted•a year ago

Motion in a Plane Physics NCERT Exemplar Solutions Class 12th Chapter Four

If $\bar{A} = (2 \hat{i} + 3 \hat{j} - \hat{k}) m a n d \vec{B} = (\hat{i} + 2 \hat{j} + 2 \hat{k}) m .$ The magnitude of component of vector $\vec{A}$ along vector $\vec{B}$ will be________m.

6 Views|Posted a year ago

Asked by Alok Kumar Singh

Answered by

Alok Kumar Singh | Contributor-Level 10

a year ago

Component of $\vec{A}$ along $\vec{B}$ = $\frac{\vec{A} . \vec{B}}{| \vec{B} |}$

= 2

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New answer posted•a year ago

Motion in a Plane Physics NCERT Exemplar Solutions Class 12th Chapter Four

If the initial velocity in horizontal direction of a projectile is unit vector $\hat{i}$ and the equation of trajectory is y = 5x(1 – x). The y component vector of the initial velocity is ________ $\hat{j}$ .

15 Views|Posted a year ago

Asked by Vishal Baghel

Answered by

Vishal Baghel | Contributor-Level 10

a year ago

y = x5 (1 - x) = x tanθ $(1 - \frac{x}{R})$

tan = 5, R = 1

$s i n θ = \frac{5}{\sqrt[]{26}}, c o s θ = \frac{1}{\sqrt[]{26}}$

$R = \frac{u^{2} s i n 2 θ}{g} = 1$

$\Rightarrow u^{2} = 26 \Rightarrow u = \sqrt[]{26} m / s$

y - component of initial velocity

= u sinθ

= $= \sqrt[]{26} \times \frac{5}{\sqrt[]{26}}$

= 5 m/s

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New answer posted•a year ago

Motion in a Plane Physics NCERT Exemplar Solutions Class 12th Chapter Four

Two projectiles thrown at $30 ° a n d 45 °$ with the horizontal respectively, reach the maximum height in same time. The ratio of their initial velocities is:

7 Views|Posted a year ago

Asked by Vishal Baghel

Answered by

Vishal Baghel | Contributor-Level 10

a year ago

H₁ = H₂

$\frac{u_{1}^{2} s i n^{2} θ_{1}}{2 g} = \frac{u_{2}^{2} s i n^{2} θ_{2}}{2 g}$

$\Rightarrow u_{1}^{2} {(s i n 30 °)}^{2} = {u_{2}}^{2} {(s i n 45 °)}^{2}$

$\Rightarrow \frac{u_{1}}{u_{2}} = \sqrt[]{2}$

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New answer posted•a year ago

Motion in a Plane Physics NCERT Exemplar Solutions Class 12th Chapter Four

Two projectiles are thrown with same initial velocity making an angle of $45 °$ and $30 °$ with the horizontal respectively. The ration of their respectively ranges will be:

4 Views|Posted a year ago

Asked by Alok Kumar Singh

Answered by

Alok Kumar Singh | Contributor-Level 10

a year ago

$R_{1} = \frac{u^{2} s i n 90}{g}$

$R_{2} = \frac{u^{2} s i n 60}{g}$

$\Rightarrow \frac{R_{1}}{R_{2}} = \frac{2}{\sqrt[]{3}}$

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New answer posted•a year ago

Motion in a Plane Physics NCERT Exemplar Solutions Class 12th Chapter Four

The trajectory of a projectile in an vertical plane is y = ax - bx², where a and b are constants and x & y are respectively the horizontal and vertical distances of the projectile from the point of projection. The angle of projection q and the maximum height attained H are respectively given by

5 Views|Posted a year ago

Asked by Payal Gupta

Answered by

Payal Gupta | Contributor-Level 10

a year ago

$y = α x - β x^{2}$

$\Rightarrow \frac{d y}{d x} = α - 2 β x = 0$

$\Rightarrow x = \frac{α}{2 β}$

$y_{m a x} = α \times \frac{α}{2 β} - β \times {(\frac{α}{2 β})}^{2} = \frac{α^{2}}{4 β}$

Range = $2 x = \frac{α}{β} = \frac{2 u^{2} s i n θ . c o s θ}{g}$

On comparing with

y = x tan θ- $\frac{g x^{2}}{2 u^{2} c o s^{2} θ}$

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New answer posted•a year ago

Motion in a Plane Physics NCERT Exemplar Solutions Class 12th Chapter Four

Starting from the origin at time $t = 0$ , with initial velocity $5 \hat{j} {m s}^{- 1}$ , a particle moves in the $x$ $- y$ plane with a constant acceleration of $(10 \hat{i} + 4 \hat{j}) {m s}^{- 2}$ . At time $t$ , its coordinates are $(20$ $m, y_{0} m)$ . The values of $t$ and $y_{0}$ are, respectively:

10 Views|Posted a year ago

Asked by Alok Kumar Singh

Answered by

Alok Kumar Singh | Contributor-Level 10

a year ago

$\overset{?}{U} = 5 \hat{j}$

$\overset{?}{a} = 10 \hat{i} + 4 \hat{j}$

$\Rightarrow \overset{?}{S} = \overset{?}{U} t + \frac{1}{2} (\overset{?}{a}) t^{2}$

$\Rightarrow 20 \hat{i} + y_{0} \hat{j} = (5 t^{2}) \hat{i} + (5 t + 2 t^{2}) \hat{j}$

$\begin{array}{r} 20 = 5 t^{2} y_{0} = 5 t + 2 t^{2} \\ t = 2 \Rightarrow 18 m . \end{array}$

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New answer posted•a year ago

Motion in a Plane Physics NCERT Exemplar Solutions Class 12th Chapter Eleven

At time t = 0 a particle starts travelling from a height $7 \hat{z} c m$ in a plane keeping z coordinate constant. At any instant of time it’s position along the $\hat{x} a n d \hat{y}$ directions are defined as 3t and 5t3 respectively. At t = 1s acceleration of the particle will be

6 Views|Posted a year ago

Asked by Vishal Baghel

Answered by

Vishal Baghel | Contributor-Level 10

a year ago

x = 3t y = 5t³

$V_{x} = \frac{d x}{d t} = 3 V_{y} = \frac{d y}{d t} = 15 t^{2}$

$a_{x} = \frac{d V_{x}}{d t} = 0 a_{y} = \frac{d V_{y}}{d t} = 30 t$

$\vec{a} = 30 t \hat{j}$

at t = 1 sec, $\vec{a} = 30 \hat{j}$

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