Question

A particle is dropped from the top of the tower it covers 40 metre in last 2 seconds find the height of the tower​

Answer 1

Answer:

45 m

Explanation:

Let the height of the tower be "H"

Total time taken to reach the ground be "T"

Then, from equation of motion

• S = (v² - u²)/(2a)

Velocity when it touches the ground is

v = √(u² + 2aS)

v = √(0² + (2 × 10 × H))

v = √(20H)

Velocity after total time "T"

v = u + at

√(20H) = 0 + gT

√(20H) = 10T

20H = 100T²

H = 100T²/20

H = 5T² ....(1)

Distance covered in "T - 2" seconds is "H - 40" metres. So, from equation of motion

• S = ut + 0.5at²

H - 40 = (0 × T) + 0.5g(T - 2)²

From equation (1)

5T² - 40 = 5(T - 2)²

From (a - b)² formula

5T² - 40 = 5[T² + 4 - 4T]

5T² - 40 = 5T² + 20 - 20T

-40 = 20 - 20T

20T = 60

T = 60/20

T = 3 seconds

Now, from equation (1)

H = 5T²

= (5 × 3²) m

= (5 × 9) m

= 45 m