Solution 8.7b
From Mechanics
(Difference between revisions)
(New page: We use the expression for the position vector obtained in part a), <math> \mathbf{r}=(8\mathbf{i}+10\mathbf{j})t+\frac{1}{2}(-10\mathbf{j}){{t}^{\ 2}} </math> From part a) the ball hits t...) |
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From part a) the ball hits the ground when <math>t=2\text{ s}</math>. | From part a) the ball hits the ground when <math>t=2\text{ s}</math>. | ||
- | The distance along the ground is the <math>\mathbf{i}</math> component giving <math>16\text{ m}</math> | + | The distance along the ground is the <math>\mathbf{i}</math> component at this time giving, |
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+ | <math>16\text{ m}</math> |
Current revision
We use the expression for the position vector obtained in part a), \displaystyle \mathbf{r}=(8\mathbf{i}+10\mathbf{j})t+\frac{1}{2}(-10\mathbf{j}){{t}^{\ 2}}
From part a) the ball hits the ground when \displaystyle t=2\text{ s}.
The distance along the ground is the \displaystyle \mathbf{i} component at this time giving,
\displaystyle 16\text{ m}