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| 183_notes:examples:finding_the_time_of_flight_of_a_projectile [2014/07/22 06:14] – pwirving | 183_notes:examples:finding_the_time_of_flight_of_a_projectile [2015/09/15 16:51] (current) – [Solution] pwirving | ||
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| Diagram of forces acting on bus once it leaves the road. | Diagram of forces acting on bus once it leaves the road. | ||
| - | {{183_notes: | + | {{183_notes: |
| General equation for position up date formula for constant force systems: | General equation for position up date formula for constant force systems: | ||
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| $$\vec{r}_{f} = \vec{r}_{i} + \vec{v}_{i} \Delta t + \dfrac{1}{2}\dfrac{\vec{F}_{net}}{m} \Delta t^2$$ | $$\vec{r}_{f} = \vec{r}_{i} + \vec{v}_{i} \Delta t + \dfrac{1}{2}\dfrac{\vec{F}_{net}}{m} \Delta t^2$$ | ||
| - | Equation for determining $\Delta{t}$ in constant force situations.<wrap todo>use vector principle; then motivate why only y matters</ | + | Equation for determining $\Delta{t}$ in constant force situations. |
| ==== Solution ==== | ==== Solution ==== | ||
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| $0 - 40 = v_{iy} \Delta{t} + \dfrac{1}{2} \dfrac{-mg}{m} \Delta{t^2}$ | $0 - 40 = v_{iy} \Delta{t} + \dfrac{1}{2} \dfrac{-mg}{m} \Delta{t^2}$ | ||
| + | The masses cancel. | ||
| $ -40 = v_{iy} \Delta{t} + -\dfrac{1}{2} {g} \Delta{t^2}$ | $ -40 = v_{iy} \Delta{t} + -\dfrac{1}{2} {g} \Delta{t^2}$ | ||
| + | |||
| + | Rearrange the equation. | ||
| $ -40 -v_{iy} \Delta{t} = -\dfrac{1}{2} {g} \Delta{t^2}$ | $ -40 -v_{iy} \Delta{t} = -\dfrac{1}{2} {g} \Delta{t^2}$ | ||
| - | $ -40 -v_{iy} | + | Sub in values for $v_{iy}$ and ${g}$ |
| - | $ 2\dfrac{{40 + v_{iy}}}{g} = \Delta{t}$ | + | $ -40 -(7 \Delta{t}) = -4.9 \Delta{t^2}$ |
| - | $ \Delta{t} = 9.59s$ | + | Solve the quadratic equation |
| + | $ \Delta{t} = 3.65s$ | ||