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Physics Motion in Plane

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2 months ago

Physics Motion in Plane Class 11th

A particle is thrown vertically upward. Its velocity at half of the maximum height is 10m/s. The maximum height attained by it is (g=10 ms^–2) :–

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alok kumar singh

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2 months ago

Physics Motion in Plane Class 11th

A ball is projected from ground at angle q with the horizontal. After t = 1 s, it is moving at 45° with the horizontal and after t = 2 second, it is moving horizontally. What is speed of projection of ball? [g = 10 m s^–2]

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alok kumar singh

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3 months ago

Physics Motion in Plane Class 11th

A ball is projected vertically up with an initial velocity. Which of the following graph represent KE of ball?

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alok kumar singh

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Graph with time must be parabola. At the highest point KE is zero

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3 months ago

Physics Motion in Plane Class 11th

Distance travelled by a particle starting form rest and moving with an acceleration of 4/3 ms^-2, in the second is

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alok kumar singh

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3 months ago

Physics Motion in Plane Class 11th

Consider a series of steps as shown. A ball is thrown from O. Find the minimum speed of directly jump to 5th step.

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alok kumar singh

Contributor-Level 10

$y = x t a n θ - \frac{g x^{2}}{2 v^{2} c o s^{2} θ}$

(2.5, 2.5) must lie on this

-> $1 = t a n θ - \frac{g * 2.5}{2 v^{2} c o s^{2} θ}$

-> $\frac{25}{2 v^{2} c o s^{2} θ} = t a n θ - 1$

-> $v^{2} = \frac{25}{2} {\frac{1 + t a n^{2} θ}{t a n θ - 1}}$

-> $v_{m i n} = 5 \sqrt[]{\sqrt[]{2} + 1}$

[Happens when tan θ = $\sqrt[]{2}$ + 1]

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3 months ago

Physics Motion in Plane Class 11th

If ratio of centripetal acceleration of two particles moving on the same path is 3: 4. Find the ratio of their tangential velocities.

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alok kumar singh

Contributor-Level 10

$a_{c} = \frac{v^{2}}{r}$ , $\frac{{(a_{c})}_{1}}{{(a_{c})}_{2}} = {(\frac{v_{1}}{v_{2}})}^{2}$

$\frac{3}{4} = {(\frac{v_{1}}{v_{2}})}^{2} \to \frac{v_{1}}{v_{2}} = \sqrt[]{3} : 2$

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3 months ago

Physics Motion in Plane Class 11th

A particle of mass m is projected with speed v at an angle of 30° with the horizontal, find its angular momentum about point of projection when it reaches its maximum height.

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alok kumar singh

Contributor-Level 10

Velocity at maximum height = vcoss30°

L = m (vcos30) H

$= m v (\frac{\sqrt[]{3}}{2}) * \frac{v^{2} s i n^{2} 30}{2 g}$

$= \sqrt[]{3} \frac{m v^{3}}{16 g}$

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3 months ago

Physics Motion in Plane Class 11th

Two vector each of magnitude A are inclined at angle q with each other, then magnitude of resultant vector is

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alok kumar singh

Contributor-Level 10

The magnitude of resultant vector (R)

$= \sqrt[]{a^{2} + b^{2} + 2 a b c o s θ}$

here a = b = A

then R = $\sqrt[]{A^{2} + A^{2} + 2 A^{2} c o s θ}$

$= A \sqrt[]{2} \sqrt[]{1 + c o s θ}$

$= \sqrt[]{2} A \sqrt[]{2 c o s^{2} \frac{θ}{2}}$

$= 2 A c o s \frac{θ}{2}$

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3 months ago

Physics Motion in Plane Class 11th

A man is swimming with speed 10 km/h in still water crosses a river of width 1 km along the shortest path in 10 minutes. The velocity of river flow is

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alok kumar singh

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3 months ago

Physics Motion in Plane Class 11th

A man wishes to swim across a river 0.5 km wide. If he can swim at the rate of 2km/h in still water and the river flows at the rate of 1km/h. The angle (with respect to the flow of the river) along which he should swim so at to reach a point exactly opposite his starting point, should be

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