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A Particle Projected with speed U at angle theta with horizontal from ground . If it is at same height from ground at time t1 and time t2 , then it's average velocity in time interval t1 to t2 is ......

Tanvir Alam , 6 Years ago
Grade 12th pass
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Eshan

Last Activity: 6 Years ago

To determine the average velocity of a particle projected at speed U at an angle θ, we need to consider the particle's motion in terms of its displacement and the time interval. When the particle reaches the same height at two different times, we can apply the formula for average velocity, which is the total displacement divided by the total time taken.

Understanding the Motion

The motion of the particle can be decomposed into horizontal and vertical components. The initial velocity can be broken down as follows:

  • Horizontal component: \( U_x = U \cdot \cos(\theta) \)
  • Vertical component: \( U_y = U \cdot \sin(\theta) \)

Vertical Motion

Due to gravity, the vertical motion of the particle can be described by the equation:

\( y(t) = U_y t - \frac{1}{2} g t^2 \),

where \( g \) is the acceleration due to gravity. When the particle reaches the same height at times \( t_1 \) and \( t_2 \), the vertical displacement for both times is equal, meaning:

\( y(t_1) = y(t_2) \).

Average Velocity Calculation

The average velocity (\( V_{avg} \)) over the time interval from \( t_1 \) to \( t_2 \) can be calculated using the formula:

\( V_{avg} = \frac{\Delta x}{\Delta t} \),

where \( \Delta x \) is the horizontal displacement and \( \Delta t \) is the time interval \( (t_2 - t_1) \).

Horizontal Displacement

The horizontal displacement during the time interval can be determined using the horizontal component of the initial velocity:

\( x(t) = U_x t = U \cos(\theta) t \).

Thus, the displacement from \( t_1 \) to \( t_2 \) is:

\( \Delta x = x(t_2) - x(t_1) = U \cos(\theta) (t_2 - t_1) \).

Putting It All Together

Substituting \( \Delta x \) into the average velocity formula gives:

\( V_{avg} = \frac{U \cos(\theta) (t_2 - t_1)}{t_2 - t_1} \).

Notice that \( (t_2 - t_1) \) cancels out, leading to:

\( V_{avg} = U \cos(\theta) \).

Final Insight

This result tells us that the average velocity of the particle over the time interval from \( t_1 \) to \( t_2 \), when it is at the same height, solely depends on the horizontal component of its initial velocity, irrespective of the vertical motion. It's a neat demonstration of how the components of motion can be analyzed separately, leading to clear insights about overall behavior. So, whenever you see a particle projected at an angle, remember that the average velocity in such scenarios can be derived simply from its horizontal motion!

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