In Weibel's, it says that:
If $F: \mathcal{A} \to \mathcal{B}$ is a right exact functor of categories, and if $A$ is projective in $\mathcal{A}$, then $L^iF(A)=0$ for all $i\ne0$.
I guess we need to show that the projective resolution is exact under the functor $F$. But I don't know how does the property of $A$ come into picture. Hope someone could help. Thanks!
If $A$ is projective, then $$ 0 \longrightarrow A \overset{\mathrm{id}_A}{\longrightarrow} A \longrightarrow 0 $$ is exact, where $\mathrm{id}_A$ is the identity morphism of $A$. Since each term is projective, it is a projective resolution for $A$. Applying $F$ gives $$ 0 \longrightarrow F(A) \longrightarrow F(A) \longrightarrow 0 \,, $$ where the map $F(A) \to F(A)$ is $F(\mathrm{id}_A) = \mathrm{id}_{F(A)}$, the identity morphism of $F(A)$. To get the $L^i F(A)$, we remove the second $F(A)$ term to get the chain complex $$ 0 \longrightarrow F(A) \longrightarrow 0 \,, $$ and take its homology. But since only the $0$-th term is nontrivial, the homology must be trivial for all $i > 0$. Note that we did not show $F$ is exact. The result of applying $F$ does end up being exact to the left of the $F(A)$ term, but only because it vanishes completely.