The formula that is useful for solving both of these problems is ...
[tex](v_2)^2-(v_1)^2=2ad\qquad\text{a=acceleration, d=distance}\\\\ \text{where $v_1$ and $v_2$ are initial and final velocities}[/tex]
9. Given v₁=15, a=9.8, d=10, find v₂.
... (v₂)² = 15² + 2·9.8·10
... v₂ = √421 ≈ 20.5 . . . . m/s
10. Given d=12 m when a=-9.8 m/s² and v₂=0, find d when a=0.17·(-9.8 m/s²).
The formula tells us that d=(v₁)²/(2a), which is to say that the distance is inversely proportional to the acceleration. If acceleration is 0.17 times that on earth, distance will be 1/0.17 ≈ 5.88 times that on earth.
(12 m)/0.17 ≈ 70.6 m
