Question

What is the minimum landing distance needed for the space shuttles orbiter if it is coming in at 98.45m/s with an average acceleration of -0.81m/s^2? Give answer in m

Answer 1

$$v^2=v_0^2+2ad,$$ $$0=v_0^2+2ad,$$ $$d=\dfrac{-v_0^2}{2a}=\dfrac{-(98.45\ \dfrac{m}{s^2})^2}{2\cdot(-0.81\ \dfrac{m}{s^2})}=5983\ m.$$