We can use equation of motion to solve this.

Using v^2 = u^2 + 2as

where, v is the final velocity, u is the initial velocity, a is the acceleration and s is the distance covered.

Here, the initial velocity (u) is zero, since the object was dropped from the tower and hence has no initial velocity. The distance traveled is the same as the height of the tower, h. The acceleration of the object is the same as the acceleration due to gravity of Earth and is equal to g (or 9.81 m/s^2).

Substituting everything in the equation, we get

v^2 = 0 + 2gh

or `v = sqrt(2gh)`

Momentum of an object is given as the product of its mass and velocity. The mass of the product is given as 500 gm or 0.5 kg and we just calculated its velocity as it touches the ground.

Thus, the momentum of the object as it touches the ground is given as:

p = m x v = `1/2 xx sqrt (2gh) = 1/2 sqrt (2gh)`

and the units are kg.m/s.

Hope this helps.

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