Solution 8.7c
From Mechanics
(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...) |
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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}
