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IF A PARTICLE IS FIRED VERTICALLY UPWARDS FROM THE SURFACE OF THE EARTH AND REACHES A HEIGHT OF 6400 KM,THE INITIAL VELOCITY OF RTHE PARTICLE IS -----------------(ASSUME R=6400KM AND G=10 M/S)

SIBASIS NAYAK , 10 Years ago
Grade 11
anser 1 Answers
Askiitians Tutor Team

Last Activity: 1 Month ago

To determine the initial velocity of a particle fired vertically upwards from the Earth's surface that reaches a height of 6400 km, we can use the principles of physics, specifically the equations of motion under the influence of gravity. Given that the radius of the Earth (R) is 6400 km and the acceleration due to gravity (g) is 10 m/s², we can derive the initial velocity needed for the particle to reach that height.

Understanding the Problem

When a particle is projected upwards, it experiences a deceleration due to gravity until it reaches its maximum height. At that point, its velocity becomes zero before it starts to fall back down. The key here is to find the initial velocity (u) required to reach a height (h) of 6400 km (or 6,400,000 meters, since we need to work in standard units).

Using the Kinematic Equation

We can use the following kinematic equation that relates initial velocity, final velocity, acceleration, and displacement:

  • v² = u² + 2as

In this equation:

  • v is the final velocity (0 m/s at the maximum height)
  • u is the initial velocity (what we want to find)
  • a is the acceleration (which will be -g, or -10 m/s², since gravity acts downwards)
  • s is the displacement (the height reached, which is 6,400,000 m)

Plugging in the Values

Since the final velocity (v) at the maximum height is 0, we can rearrange the equation:

  • 0 = u² + 2(-10)(6400000)

This simplifies to:

  • 0 = u² - 128000000

Now, we can solve for u²:

  • u² = 128000000

Taking the square root gives us:

  • u = √128000000

Calculating the Initial Velocity

Now, let's calculate the square root:

  • u ≈ 11314.0 m/s

Final Thoughts

The initial velocity required for the particle to reach a height of 6400 km is approximately 11314 m/s. This calculation assumes a simplified model where air resistance is neglected and gravity remains constant throughout the ascent, which is a reasonable approximation for this height relative to the Earth's radius.

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