Solution 8.7a
From Mechanics
Ian (Talk | contribs)
(New page: The ball hits the ground when the <math>\mathbf{j}</math> component of the position vector is zero. We calculate the position vector using <math> \mathbf{r}=\mathbf{u}t+\frac{1}{2}\math...)
Next diff →
Revision as of 11:36, 18 April 2010
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{i}\ and \displaystyle \ \mathbf{r}_{0}=0 giving,
\displaystyle \mathbf{r}=(8\mathbf{i}+10\mathbf{j})t+\frac{1}{2}(-10\mathbf{i}){{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.
