The question is: A body of mass 10 g falls from a height of 3 m into a pile of sand. The body penetrates the sand at distance of 3 cm before stopping. What force has the sand exerted on the body?
The solution is: Let $v$ be the velocity of the body at the instant it reaches the pile of sand Then from the relation $v=v_0^2+2 g y$, we have $$ \begin{aligned} v^{2} &=0+2 \times\left(9.8 \text {m/s}^2\right) \times 3 \,\text{m:} \\ &=58.8\,(\text{m/s})^{2} \end{aligned} $$ This velocity is reduced to zero due to the deceleration ' $a$ ' produced by the sand. Thus, from the relation $u^{2}=v_{0}^{2}+2 a y$, we have $$ \begin{array}{l} 0=58.8+2 a(0.03 \text {m}) \\ a=-\frac{58.8}{2 \times 0.03}=-980\,\text {m/s}^{2} \end{array} $$ The mass of the body is $10 \mathrm{g}=0.01\, \mathrm{kg}$. Hence the (retarding) force exerted by the sand on it is $$ \begin{aligned} F &=m a \\ &=0.01\,\mathrm{kg} \times\left(-980\,\mathrm{m} / \mathrm{s}^2\right) \\ &=-9.8\,\mathrm{N} \end{aligned} $$
Now My question is if the answer is correct cause when the body will reach on the surface the gravity will still work. So, do I need to add 9.8 m/s^2 in the retardation.