# a stone is dropped from the height 24.5 it hits the ground and its 3 centimetre into it what is the retardation produced by the ground

As it's not exactly specified in question, assuming that the stone stops after it's 3 cm into the ground.

Considering the accleration due to gravity, g, as 10m/s

^{2}

Time taken to reach the ground, using the equation of motion, 2 a s = v

^{2}- u

^{2}

=> 2 * 10 * 24.5 = v

^{2}- 0

=> v

^{2}= 490

Now, we know that when it hits the ground, the stone travels 3cm = 0.03m and then stops. It's velocity when it hits the ground is $\sqrt{490}$ m/s, and it's final velocity in this case is 0 m/s.

Then, finding retardation using the same equation, 2 a s = v

^{2}- u

^{2}

=> 2*a*0.03 = 0 - 490

=> a = - $\frac{490}{2*0.03}$

=> a = - 8166.667 m/s

^{2}

(the negative sign stands for retardation).

Regards

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