Solution 8.7a
From Mechanics
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We calculate the position vector using <math> \mathbf{r}=\mathbf{u}t+\frac{1}{2}\mathbf{a}{{t}^{\ 2}}+{{\mathbf{r}}_{0}} </math>. | We calculate the position vector using <math> \mathbf{r}=\mathbf{u}t+\frac{1}{2}\mathbf{a}{{t}^{\ 2}}+{{\mathbf{r}}_{0}} </math>. | ||
| - | Here <math>\mathbf{u}=8\mathbf{i}+10\mathbf{j}\text{ m}{{\text{s}}^{\text{-1}}}</math>, <math>\ \mathbf{a}=-10\mathbf{j}\ </math> and <math>\ \mathbf{r}_{0}=0</math> giving, | + | Here <math>\mathbf{u}=8\mathbf{i}+10\mathbf{j}\text{ m}{{\text{s}}^{\text{-1}}}</math>, <math>\ \mathbf{a}=-10\mathbf{j}\ \text{ m}{{\text{s}}^{\text{-2}}}</math> and <math>\ \mathbf{r}_{0}=0</math> giving, |
<math> \mathbf{r}=(8\mathbf{i}+10\mathbf{j})t+\frac{1}{2}(-10\mathbf{j}){{t}^{\ 2}} </math> | <math> \mathbf{r}=(8\mathbf{i}+10\mathbf{j})t+\frac{1}{2}(-10\mathbf{j}){{t}^{\ 2}} </math> | ||
Current revision
The ball hits the ground when the \displaystyle \mathbf{j} component of the position vector is zero.
We calculate the position vector using \displaystyle \mathbf{r}=\mathbf{u}t+\frac{1}{2}\mathbf{a}{{t}^{\ 2}}+{{\mathbf{r}}_{0}} .
Here \displaystyle \mathbf{u}=8\mathbf{i}+10\mathbf{j}\text{ m}{{\text{s}}^{\text{-1}}}, \displaystyle \ \mathbf{a}=-10\mathbf{j}\ \text{ m}{{\text{s}}^{\text{-2}}} and \displaystyle \ \mathbf{r}_{0}=0 giving,
\displaystyle \mathbf{r}=(8\mathbf{i}+10\mathbf{j})t+\frac{1}{2}(-10\mathbf{j}){{t}^{\ 2}}
The \displaystyle \mathbf{j} component of this position vector is \displaystyle 10t-5{{t}^{\ 2}}.
This is zero if \displaystyle t=2\ \text{s}.
The other solution \displaystyle t=0 is the trivial solution restating the fact that the ball is at the origin at the start.
