From the top of a tower of height ‘H’, a body is thrown vertically upwards with a speed ‘u’. Time taken by the body to reach the ground is ‘3’ times the time taken by it to reach the highest point in its path. Then, the speed u is

• Height of the tower: H
• Initial upward speed: u
• Total time taken: t_total = 3t_up
• t_up: Time taken to reach the highest point.

Key Physics Concepts:

1. At the highest point, the velocity of the body becomes zero. Using the kinematic equation for vertical motion:

   • v = u - g t_up

   At the highest point, v = 0, so:

   t_up = u / g

2. Total time taken to reach the ground is given as:

   t_total = 3 t_up = 3 × u / g

Step 1: Time Taken for the Downward Journey

The total time can be split into:

t_total = t_up + t_down

Substituting t_total = 3t_up:

3t_up = t_up + t_down

Simplifying:

t_down = 2t_up

Step 2: Distance Covered During Downward Motion

The body first reaches the highest point (displacement = H + u²/2g) and then falls back to the ground. The total displacement downward includes the height of the tower H and the height it rose above the tower.

Using the kinematic equation for downward motion: H + u²/2g = (1/2) g t_down²

Substitute t_down = 2t_up = 2u / g: H + u²/2g = (1/2) g (2u / g)²

Simplify:

H + u²/2g = (1/2) g × (4u² / g²)

H + u²/2g = 2u² / g

Rearrange:

H = (2u² / g) - (u² / 2g)

H = (4u² - u²) / 2g

H = 3u² / 2g

Step 3: Solving for u

Rearrange the equation to find u:

u² = (2gH) / 3

u = √(2gH / 3)

Final Answer:

The speed u is:

u = √(2gH / 3)