The formula that is useful for solving both of these problems is ...
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
Consider right triangle ΔABC with legs AC and BC and hypotenuse AB. Draw the altitude CD.
1. Theorem: The length of each leg of a right triangle is the geometric mean of the length of the hypotenuse and the length of the segment of the hypotenuse adjacent to that leg.
According to this theorem,
Let BC=x cm, then AD=BC=x cm and BD=AB-AD=3-x cm. Then
Take positive value x. You get
2. According to the previous theorem,
Then
Answer:
This solution doesn't need CD=2 cm. Note that if AB=3cm and CD=2cm, then
This means that you cannot find solutions of this equation. Then CD≠2 cm.