Solution 8.7b
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(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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Revision as of 11:42, 18 April 2010
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 giving \displaystyle 16\text{ m}
