Question

a particle is dropped from the top of a very high tower then the distance covered in the 5th second of its fall is

Answer 1

Answer:

Distance covered in 5th sec = 45 m

Explanation:

a particle is dropped from the top of a very high tower then the distance covered in the 5th second of its fall is

S = ut + (1/2)at²

u = 0 m/s dropped

a = g = 10 m/s²

t = time

Distance covered in 5 sec

S = 0 + (1/2)10 * 5² = 125 m

Distance covered in 4 sec

S = 0 + (1/2)10 * 4² = 80 m

Distance covered in 5th sec = 125 - 80 = 45 m

or

Speed after 4 sec

v = u + at = 0 + 10*4 = 40 m/s

speed after 5 sec = 0 + 10*5 = 40 m/s

v² - u² = 2as

=> 50² - 40² = 2 * 10 * S

=> 2500 - 1600 = 20S

=> 900 = 20S

=> S = 45 m

Answer 2

Answer:

Distance covered in 5th sec = 45 m

Explanation:

a particle is dropped from the top of a very high tower then the distance covered in the 5th second of its fall is

S = ut + (1/2)at²

u = 0 m/s dropped

a = g = 10 m/s²

t = time

Distance covered in 5 sec

S = 0 + (1/2)10 * 5² = 125 m

Distance covered in 4 sec

S = 0 + (1/2)10 * 4² = 80 m

Distance covered in 5th sec = 125 - 80 = 45 m

or

Speed after 4 sec

v = u + at = 0 + 10*4 = 40 m/s

speed after 5 sec = 0 + 10*5 = 40 m/s

v² - u² = 2as

=> 50² - 40² = 2 * 10 * S

=> 2500 - 1600 = 20S

=> 900 = 20S

=> S = 45 m