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A ball is projected with 20sqrt2 m//s at...

A ball is projected with `20sqrt2 m//s` at angle `45^(@)` with horizontal.The angular velocity of the particle at highest point of its journey about point of projection is

A

1 rad/s

B

0.2 rad/s

C

0.3 rad/s

D

0.4 rad/s

Text Solution

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The correct Answer is:
To find the angular velocity of the ball at the highest point of its journey about the point of projection, we can follow these steps: ### Step 1: Determine the initial velocity components The ball is projected with an initial velocity \( u = 20\sqrt{2} \, \text{m/s} \) at an angle \( \theta = 45^\circ \) with the horizontal. We can find the horizontal and vertical components of the initial velocity. \[ u_x = u \cos \theta = 20\sqrt{2} \cos 45^\circ = 20\sqrt{2} \cdot \frac{1}{\sqrt{2}} = 20 \, \text{m/s} \] \[ u_y = u \sin \theta = 20\sqrt{2} \sin 45^\circ = 20\sqrt{2} \cdot \frac{1}{\sqrt{2}} = 20 \, \text{m/s} \] ### Step 2: Find the velocity at the highest point At the highest point of its trajectory, the vertical component of the velocity becomes zero, and only the horizontal component remains. \[ v = u_x = 20 \, \text{m/s} \] ### Step 3: Calculate the maximum height The maximum height \( H \) can be calculated using the formula: \[ H = \frac{u_y^2}{2g} \] Substituting \( u_y = 20 \, \text{m/s} \) and \( g = 10 \, \text{m/s}^2 \): \[ H = \frac{(20)^2}{2 \cdot 10} = \frac{400}{20} = 20 \, \text{m} \] ### Step 4: Determine the radius of rotation At the highest point, the radius \( R \) from the point of projection to the ball is equal to the horizontal distance traveled when the ball reaches the maximum height. Since the ball is projected at \( 45^\circ \), the horizontal range \( R \) can be calculated as: \[ R = \frac{u^2 \sin 2\theta}{g} = \frac{(20\sqrt{2})^2 \sin 90^\circ}{10} = \frac{800}{10} = 80 \, \text{m} \] At the highest point, the horizontal distance from the point of projection is \( R/2 = 40 \, \text{m} \). ### Step 5: Calculate the angular velocity The angular velocity \( \omega \) can be calculated using the formula: \[ \omega = \frac{v}{R} \] Substituting \( v = 20 \, \text{m/s} \) and \( R = 40 \, \text{m} \): \[ \omega = \frac{20}{40} = 0.5 \, \text{rad/s} \] ### Final Answer The angular velocity of the particle at the highest point of its journey about the point of projection is \( 0.5 \, \text{rad/s} \). ---
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