A particle is dropped from the top of a high tower,then the distance covered in the fifth second of its fall is?

Question

Answer

Distance Covered in the Fifth Second of the Fall

Given Data: The particle is dropped from the top of a high tower.

Explanation:

First Second: In the first second, the particle covers a certain distance due to gravity.
Second Second: In the second second, the particle covers a greater distance as it gains speed.
Third Second: The distance covered in the third second is even greater than the second second.
Fourth Second: Similar to the previous seconds, the particle covers more distance in the fourth second.
Fifth Second: By the fifth second, the particle has gained considerable speed and covers the maximum distance in this second.

Calculating the Distance Covered in the Fifth Second:

Formulas: The distance covered by a falling object can be calculated using the formula: d = (1/2) * g * t^2, where d is the distance, g is the acceleration due to gravity, and t is the time.
Acceleration due to Gravity: The acceleration due to gravity is approximately 9.8 m/s^2.
Time: As the particle is in its fifth second of fall, the time taken is 5 seconds.
Calculate: Substituting the values in the formula, we get: d = (1/2) * 9.8 * 5^2 = 122.5 meters.

Conclusion:

Therefore, the distance covered by the particle in the fifth second of its fall is 122.5 meters.