Solution 8.7c

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(New page: The maximum height is when the <math>\mathbf{j}</math> component of the velocity is zero. Using <math>\mathbf{v}=\mathbf{u}+\mathbf{a}t \ </math> with <math>\mathbf{u}=8\mathbf{i}+10\mat...)
Current revision (12:28, 18 April 2010) (edit) (undo)
 
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Thus the ball is at its maximum height when <math>t=1\text{ s}</math>.
Thus the ball is at its maximum height when <math>t=1\text{ s}</math>.
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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>.
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The maximum height is the <math>\mathbf{j}</math> component of this vector at <math>t=1\text{ s}</math>
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<math>10\times 1+\frac{1}{2}\left( -10 \right)\times {{1}^{2}}=5\ \text{m}</math>

Current revision

The maximum height is when the \displaystyle \mathbf{j} component of the velocity is zero.

Using \displaystyle \mathbf{v}=\mathbf{u}+\mathbf{a}t \ with \displaystyle \mathbf{u}=8\mathbf{i}+10\mathbf{j}\text{ m}{{\text{s}}^{\text{-1}}} and \displaystyle \ \mathbf{a}=-10\mathbf{j}\ \text{ m}{{\text{s}}^{\text{-2}}} ,

\displaystyle \mathbf{v}=8\mathbf{i}+10\mathbf{j}+(-10\mathbf{j})t \

Thus the ball is at its maximum height when \displaystyle t=1\text{ s}.

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}} .

The maximum height is the \displaystyle \mathbf{j} component of this vector at \displaystyle t=1\text{ s}

\displaystyle 10\times 1+\frac{1}{2}\left( -10 \right)\times {{1}^{2}}=5\ \text{m}