Question

a body projected horizontally with a velocity u from the top of a tower strikes the ground with a velocity 2u the height of the tower is​

Answer 1

3u*u/(2g)

Explanation:

square of vertical velocity at ground is 3u*u as horizontal velocity wont change

Answer 2

The height of the tower is 3u² / 2g.

Given: A body projected horizontally with a velocity 'u' from the top of a tower strikes the ground with a velocity '2u'.

To Find: The height of the tower.

Solution:

We can solve the numerical by using the formula of motion, which states that;

         v² = u² + 2aS                                                             ....(1)

Where v = final velocity, u = initial velocity, a = acceleration, S = Height.

Coming to the numerical, we are given that;

The initial velocity (u) = u

The final velocity (v) = 2u

The acceleration (a) = g = 9.8 m/s²

Let the height be 'S'.

Putting respective values in (1), we get;

              v² = u² + 2aS        

         ⇒ ( 2u )² = u² + 2gS

         ⇒  4u² = u² + 2g S

         ⇒  2gS = 3u²

         ⇒  S = 3u² / 2g

Hence, the height of the tower is 3u² / 2g.

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