Question

a particle fall from the distance of 122.5m freely under gravity what is the time and final velocity of the particle

Answer 1

t = sqrt(2H/g)
= sqrt(2 × 122.5 / 9.8)
= 5 seconds

v = sqrt(2Hg)
= sqrt(2 × 122.5 × 9.8)
= 49 m/s

Time taken to fall: 5 seconds
Final velocity of particle: 49 m/s

Answer 2

Concept:

We can calculate the time period of a particle using the formula,

t = √2H/g where H is the distance of the particle falling under gravity and g is the acceleration due to gravity.

Given:

The distance of particle falling under gravity is, g = 22.5 m

Find:

We need to determine:

(a) The time of the particle

(b) The final velocity of the particle

Solution:

We can calculate the time period of a particle using the formula,

t = √2H/g where H is the distance of the particle falling under gravity and g is the acceleration due to gravity.

Therefore, t = √2(122.5)/ 9.8, Where g = 9.8

Therefore, t = √2(122.5)/ 9.8

t = 5 seconds.

Now, the final velocity becomes, v = √2gH

Therefore, velocity, v = √2×9.8×122.5

v = 49 m/s

Thus, The time and final velocity of the particle are 5 seconds and 49 m/s respectively.

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