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Average velocity of a particle in projec...

Average velocity of a particle in projectile motion between its starting point and the highest point of its trajectory is (projectin speed = u, angle projection from horizontal =`theta`)

A

`u cos theta`

B

`u/2 (sqrt(1+3cos^2theta))`

C

`u/2(sqrt(2+cos^2theta))`

D

`u/2 (sqrt(1+cos^2theta))`

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AI Generated Solution

The correct Answer is:
To find the average velocity of a particle in projectile motion between its starting point and the highest point of its trajectory, we can follow these steps: ### Step-by-Step Solution: 1. **Identify the Initial Parameters**: - Let the initial speed of the projectile be \( u \). - The angle of projection from the horizontal is \( \theta \). 2. **Determine the Components of Velocity**: - The horizontal component of the initial velocity is given by: \[ u_x = u \cos \theta \] - The vertical component of the initial velocity is given by: \[ u_y = u \sin \theta \] 3. **Calculate the Time to Reach Maximum Height**: - The time taken to reach the maximum height (let's denote it as \( t \)) can be calculated using the formula: \[ t = \frac{u_y}{g} = \frac{u \sin \theta}{g} \] - Since we are interested in the time to reach the highest point, we can denote this as \( t = \frac{T}{2} \), where \( T \) is the total time of flight. 4. **Determine the Maximum Height (h)**: - The maximum height reached by the projectile is given by: \[ h = \frac{u^2 \sin^2 \theta}{2g} \] 5. **Calculate the Displacement**: - The displacement from the starting point to the highest point is purely vertical, which is \( h \). 6. **Calculate the Average Velocity**: - The average velocity (\( V_{avg} \)) is defined as the total displacement divided by the total time taken to reach that displacement: \[ V_{avg} = \frac{\text{Displacement}}{\text{Time}} = \frac{h}{t} \] - Substituting the values we have: \[ V_{avg} = \frac{\frac{u^2 \sin^2 \theta}{2g}}{\frac{u \sin \theta}{g}} = \frac{u^2 \sin^2 \theta}{2g} \cdot \frac{g}{u \sin \theta} \] - Simplifying this gives: \[ V_{avg} = \frac{u \sin \theta}{2} \] 7. **Final Expression**: - The average velocity of the particle in projectile motion between its starting point and the highest point of its trajectory is: \[ V_{avg} = \frac{u}{2} \sqrt{1 + 3 \cos^2 \theta} \]

To find the average velocity of a particle in projectile motion between its starting point and the highest point of its trajectory, we can follow these steps: ### Step-by-Step Solution: 1. **Identify the Initial Parameters**: - Let the initial speed of the projectile be \( u \). - The angle of projection from the horizontal is \( \theta \). ...
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DC PANDEY ENGLISH-PROJECTILE MOTION-Level - 1 Single Correct
  1. Identify the correct statement related to the projectile motion.

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  5. A particular has initial velocity , v=3hati+4hatj and a constant force...

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  6. A body is projected at an angle 60^@ with horizontal with kinetic ener...

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  7. If T1 and T2 are the times of flight for two complementary angles, the...

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  10. Two stones are projected with the same speed but making different angl...

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  11. A ball is projected upwards from the top of a tower with a velocity 50...

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  12. Average velocity of a particle in projectile motion between its starti...

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  13. A train is moving on a track at 30ms^-1. A ball is thrown from it perp...

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  14. A body is projected at time t = 0 from a certain point on a planet's s...

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  15. A particle is fired horizontally form an inclined plane of inclination...

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  16. A fixed mortar fires a bomb at an angle of 53^@ above the horizontal w...

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