1. **State the problem:** Calculate the height $S(t)$ of an object thrown upwards with initial velocity $v_0=28$ m/s from an initial height $h_0=3$ m after $t=2$ seconds, under gravity $g=9.8$ m/s$^2$. 2. **Formula:** The height as a function of time is given by $$S(t) = -\frac{1}{2} g t^2 + v_0 t + h_0$$ where $-\frac{1}{2} g t^2$ accounts for the downward acceleration due to gravity. 3. **Substitute values:** $$S(2) = -\frac{1}{2} \times 9.8 \times 2^2 + 28 \times 2 + 3$$ 4. **Calculate each term:** $$-\frac{1}{2} \times 9.8 \times 4 = -\frac{1}{2} \times 39.2 = -19.6$$ 5. **Sum all terms:** $$S(2) = -19.6 + 56 + 3$$ 6. **Simplify:** $$S(2) = (-19.6 + 56) + 3 = 36.4 + 3 = 39.4$$ 7. **Final answer:** The height after 2 seconds is $\boxed{39.4}$ meters. **Note:** Your original calculation had an error in summing the terms; the correct height is 39.4 m, not 20.3 m.