Physics Motion in Plane

A projectile travels at θ below horizontal 1 sec after projection as shown. What is the value of θ?

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Physics Motion in Plane Motion in a Plane

Contributor-Level 10

$t a n θ = \frac{v_{y}}{v_{x}} = \frac{u_{y} - g t}{u_{x}}$

$\Rightarrow t a n θ = | \frac{8 - 10}{4} | = | - \frac{1}{2} |$

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3 weeks ago

Two particles start moving from the same point in two different directions. The first particle moves along the positive x-axis with a constant acceleration of 3 m/s². The second particle moves in a straight line making an angle of 60° with the positive x-axis with a constant speed of 24 m/s. Find the time after which the relative velocity of the particles is minimum.

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Physics Motion in Plane Motion in a Plane

Contributor-Level 10

v? = 3tî v? = 24cos 60°î + 24sin 60°? = 12î + 12√3?
v? = v? – v? = (12 – 3t)î + 12√3?
It is minimum when 12 - 3t = 0 ⇒ t = 4sec

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3 weeks ago

A particle is moving in a circle of radius 1m with angular velocity (ω) given as a function of angular displacement ( θ ) by the relation ω = θ² + 2θ. The total acceleration of the particle when θ = 1rad is:

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Contributor-Level 10

ω = θ² + 2θ
α = (ωdω)/dθ = (θ² + 2θ) (2θ + 2)
At θ = 1rad.
ω = 3rad/s and α = 12rad/s²
a? = αR = 12 m/s² a? = ω²R = 9 m/s² A? = √ (a? ² + a? ²) = 15 m/s²

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3 weeks ago

Which one of the following relations is true for two unit vector $\hat{A} a n d \hat{B}$ making an angle to each other?

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Contributor-Level 9

$| \hat{A} + \hat{B} | = \sqrt[]{A^{2} + B^{2} + 2 A B c o s θ}$

$= \sqrt[]{1 + 1 + 2 c o s θ}$

$= \sqrt[]{2 (1 + c o s θ)}$

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

$| \hat{A} + \hat{B} | = 2 c o s \frac{θ}{2}$ -(1)

$| \hat{A} - \hat{B} | = \sqrt[]{A^{2} + B^{2} - 2 A B c o s θ}$

$= \sqrt[]{1 + 1 - 2 c o s θ}$

$= 2 s i n \frac{θ}{2}$ -(2)

(2) $\div$ (1)

$\frac{| \hat{A} - \hat{B} |}{| \hat{A} + \hat{B} |} = \frac{2 s i n \frac{θ}{2}}{2 c o s \frac{θ}{2}}$

$\Rightarrow | \hat{A} - \hat{B} | = | \hat{A} + \hat{B} | t a n \frac{θ}{2}$

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3 weeks ago

$\vec{A}$ is a vector quantity such that $| \vec{A} |$ = non-zero constant. Which of the following expression is true for $\vec{A}$ ?

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Contributor-Level 9

$\vec{A} * \vec{A} = A A s i n θ \hat{n}$

$= A A s i n 0 ° \hat{n}$

= 0 [since Angle between the vectors are zero degree]

$\vec{A} * \vec{A} = 0$

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3 weeks ago

A grass hopper can jump maximum distance 1.6 m. It spends negligible time on ground. How far can it go in 2√2s (in meters)?

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Physics Motion in Plane Motion in a Plane

Contributor-Level 10

Maximum range is obtained at 45?
u²/g = 1.6 or u = 4m/s
T = (2u sin 45? )/g = (2*4* (1/√2)/10 = 0.4√2s
Number of jumps in a given time, n = t/T = 2√2 / 0.4√2 = 5

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3 weeks ago

A flying disc of radius r is moving with constant speed v? while rotating anticlockwise with angular speed v?/r along a curve PQ as shown below. Radius of curvature of curve at the instant is 4r. The magnitude of acceleration of point A at the instant is (3v₀²)/(nr). Find n.

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Contributor-Level 10

a? = v? ²/4r
a_A? = (v? ²/r²) * r = v? ²/r
a_A = 3v? ²/4r

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3 weeks ago

A particle moving in a circle of radius R with a uniform speed takes a time T to complete one revolution. If this particle were projected with the same speed at an angle ' θ ' to the horizontal, the maximum height attained by it equals 4R. The angle of projection. θ, is then given by:

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alok kumar singh

Contributor-Level 10

T = 2πR/v ⇒ v = 2πR/T
H? = (v²sin²θ)/2g = (2π²R²sinθ)/ (gT²) = 4R
sinθ = (2gT²)/ (π²R)¹/²
θ = sin? ¹ [ (2gT²)/ (π²R)]¹/²

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3 weeks ago

The angular acceleration of a body, moving along the circumference of a circle, is:

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Contributor-Level 9

The angular acceleration direction is given along angular velocity or opposite to angular velocity depending upon whether angular velocity magnitude is increasing or decreasing and this direction remains along the axis of circular motion.

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3 weeks ago

A bullet is fired from a gun at the speed of 280 m/s in the direction $30^{?}$ above the horizontal. The maximum height attained by the bullet is $(g = 9.8 {m s}^{- 2}, s i n ? 30^{?} = 0.5)$

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