Find distance through which the body will penetrates:
A body having mass 25kg falls on the ground from height of 19.6m. The body penetrates into ground. Find distance through which the body will penetrates into ground, if resistance by the ground to penetrate is constant and equal to 4998N. Take g = 9.8m/sec^{2}.
Sol : Given:
m = 25Kg, h = 19.6m, s = ?, F_{r} = 4998N, g = 9.8m/sec^{2}
Let us consider the motion of the body from height of 19.6m to the ground surface,
Initial velocity = u = 0,
Let final velocity of body when it reaches to the ground = v, Using equation, v^{2} = u^{2} + 2gh
v^{2} = (0)2 + 2 X 9.8 X 19.6
v = 19.6m/sec ...(i)
When body is penetrating in to the ground, resistance to penetration is acting in the upward direction. (Resistance is acting in the opposite direction of motion of body.) But weight of body is acting in downward direction.
Weight of the body = mg = 25 X 9.8 = 245N ...(ii)
Upward resistance to penetrate = 4998N
Net force acting in upward direction = F
F = F_{r }- mg = 4998 - 245 = 4753N ...(iii)
By using F = ma, 4753 = 25 X a
a=190.12m/sec^{2} ...(iv)
Now, calculation for distance to penetrate
Consider motion of the body from the ground to the point of penetration in to ground. Let distance of penetration = s,
Final velocity = v,
Initial velocity = u = 19.6m/sec, Retardation a = 190.12m/sec2
By using relation, v^{2} = u^{2} - 2as
(0)^{2} = (19.6)^{2} - 2 X 190.12 X S
S = 1.01m .......ANS