Question

pls solve this i will mark it as brainliest​

Answer 1

When the particle is thrown vertically upwards,

→ initial velocity = u

→ acceleration = - g

When the particle is thrown upwards, it travels some distance upwards, say x, and when it falls down then, the particle will travel the same distance x, and then travels the distance equal to the height of the tower to reach the ground.

Therefore, displacement of the particle to reach the ground = x - x - H = - H

Let the time taken by the particle to reach the ground be T and that to reach the highest point be t.

Its given that T = tn.

By second kinematic equation, the displacement of the particle when it reaches the highest point is,

x = u t - g t² / 2

And, the displacement of the particle when it reaches the ground is,

- H = u T - g T² / 2

Since T = t n,

- H = u (t n) - g (t n)² / 2

- H = u t n - g t² n² / 2 → (1)

Since the velocity at the highest point is 0, by first kinematic equation,

u - gt = 0

t = u / g

Then (1) becomes,

- H = u (u / g) n - g (u / g)² n² / 2

- H = u² n / g - u² n² / 2g

H = u² n [n / 2 - 1] / g

gH = u² n [n - 2] / 2

2gH = n u² (n - 2)

Hence (a) is the answer.