Fewpal
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A body of mass 25kg falls on the ground from a height of 19.6m. The body penetrates into the ground. Find the distance through which the body will penetrates into the ground, if the resistance by the ground to penetrate is constant and equal to 4998N. Take g = 9.8m/sec2.

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Given that: 

m = 25Kg, h = 19.6m, s = ?, Fr = 4998N, g = 9.8m/sec2

Let us first consider the motion of the body from a height of 19.6m to the ground surface, 

Initial velocity = u = 0, 

Let final velocity of the body when it reaches to the ground = v, 

Using the equation, v2 = u2 + 2gh

v2 = (0)2 + 2 X 9.8 X 19.6

v = 19.6m/sec ...(i)

When the body is penetrating in to the ground, the resistance to penetration is acting in the upward direction. (Resistance is always acting in the opposite direction of motion of body.) But the weight of the body is acting in the downward direction. 

Weight of the body = mg = 25 X 9.8 = 245N ...(ii) 

Upward resistance to penetrate = 4998N Net force acting in the upward direction = F 

F = Fr – mg = 4998 – 245 = 4753N ...(iii) 

Using F = ma, 4753 = 25 X a 

a = 190.12 m/sec2 ...(iv) 

Now, calculation for distance to penetrate Consider the motion of the body from the ground to the point of penetration in to ground.

Let the distance of penetration = s,

Final velocity = v

Initial velocity = u = 19.6m/sec,

Retardation a = 190.12m/sec2

Using the relation, v2 = u2 – 2as

(0)2 = (19.6)2 – 2 X 190.12 X S

S = 1.01m.

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