a body falling from rest has a velocity V after it falls through a height h .the distance it has to fall down for its velocity to become double is

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￼Madhav Jindal asked in PhysicsA body freely falling from rest has a velocity v after it falls through a distance h. The distance it has to fall down further, for its velocity to become double, is ... times h 4 Follow0￼Jaya Mukkavilli answered this37 helpful votes in Physics, Class XII-ScienceDear student,The body is falling freely under gravity, so the gravitational acceleration = a = gThe initial velocity of the body = u = 0After covering distance = hThe velocity becomes = vSo, v2- u2 = 2ghor v2 = 2ghAfter sometime (t) its velocity becomes double = v' = 2vLet the distance covered in time (t) = h'Here initial velocity = vSo, v'2 - v2 = 2gh'(2v)2- v2 = 2gh'4v2-v2 = 2gh'3v2  = 2gh'3(2gh) = 2gh'3h = h'Thus h' is 3 times h.

Answer

v^2=2gh, when v becomes 2v ;(2v)^2=2gH ;so 4v^2=2gH then 4(2gh)=2gH. therefore H=4h solved

Answer

Explanation:

To determine the distance a body must fall for its velocity to become double, we can use the laws of motion and equations of motion.

Key Concepts:

1. Velocity: Velocity is a vector quantity that measures the rate at which an object changes its position. It is defined as the rate of change of displacement with respect to time. Velocity is given by the equation v = u + at, where v is the final velocity, u is the initial velocity, a is the acceleration, and t is the time taken.

2. Acceleration due to gravity: When an object falls freely under the influence of gravity, it experiences a constant acceleration called the acceleration due to gravity. On the surface of the Earth, this acceleration is approximately 9.8 m/s² and is denoted by the symbol 'g'.

3. Equations of motion: The equations of motion describe the relationship between displacement, velocity, acceleration, and time. For a body falling freely under gravity, the relevant equation is given by the formula v² = u² + 2gh, where v is the final velocity, u is the initial velocity (which is zero in this case as the body starts from rest), g is the acceleration due to gravity, and h is the height or distance fallen.

Calculating the distance:

1. Let's assume the initial velocity of the body is zero (u = 0) as it falls from rest.

2. The final velocity (v) after falling through a height (h) can be determined using the equation v² = u² + 2gh. As the body starts from rest, the equation simplifies to v² = 2gh.

3. We are given that the velocity becomes double, so the final velocity becomes 2v.

4. Substituting the values in the equation, we have (2v)² = 2gh. Simplifying further, we get 4v² = 2gh.

5. Dividing both sides of the equation by 4, we obtain v² = 0.5gh.

6. Rearranging the equation, we get h = (v²) / (2g).

7. To find the distance the body must fall for its velocity to become double, we substitute the value of v (initial velocity) into the equation.

8. The distance is given by the formula h = (V²) / (2g).

Thus, the distance the body has to fall down for its velocity to become double is (V²) / (2g).

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