Question

A projectile is launched from the point $\mathbf{(}\mathit{x}\mathbf{}\mathbf{=}\mathbf{}\mathbf{0}\mathbf{,}\mathit{y}\mathbf{}\mathbf{=}\mathbf{}\mathbf{0}\mathbf{)}$, with velocity$\mathbf{(}\mathbf{12}\mathbf{.}\mathbf{0}\hat{\mathbf{i}}\mathbf{+}\mathbf{49}\mathbf{.}\mathbf{0}\hat{\mathbf{j}}\mathbf{)}\mathbf{}\mathbf{m}\mathbf{/}\mathbf{s}$, at$\mathit{t}\mathbf{=}\mathbf{0}$. (a) Make a table listing the projectile's distance$\mathbf{\left|}\overrightarrow{\mathbf{r}}\mathbf{\right|}$from the origin at the end of each second thereafter, for$\mathbf{0}\mathbf{\le }\mathit{t}\mathbf{\le }\mathbf{10}\mathbf{s}$. Tabulating the$\mathit{x}$and$\mathit{y}$coordinates and the components of velocity${\mathit{v}}_{\mathbf{x}}$and${\mathit{v}}_{\mathbf{y}}$will also be useful. (b) Notice that the projectile's distance from its starting point increases with time, goes through a maximum, and starts to decrease. Prove that the distance is a maximum when the position vector is perpendicular to the velocity. Suggestion: Argue that if$\overrightarrow{\mathbf{v}}$is not perpendicular to $\overrightarrow{\mathbf{r}}$, then$\mathbf{\left|}\overrightarrow{\mathbf{r}}\mathbf{\right|}$must be increasing or decreasing. (c) Determine the magnitude of the maximum displacement. (d) Explain your method for solving part (c).

Short answer

a) The table of the values ofis shown in the table below

t(sec)	Vec(r)
0	0
1	45.7
2	82.0
3	109
4	127
5	136
6	138
7	133
8	124
9	117
10	120

b) The distance is maximum when the position vector is perpendicular to the velocity vector.

c) The maximum distance is$1.38\text{m}$.

d) The maxima and minima condition is used to calculate the maximum magnitude of the position vector.

Step-by-step solution

x coordinate of the projectile.

Formula to calculate thecoordinate of the projectile

$\mathit{x}\mathbf{\left(}\mathit{t}\mathbf{\right)}\mathbf{=}{\mathit{u}}_{\mathbf{x}}\mathit{t}\mathbf{+}\frac{\mathbf{1}}{\mathbf{2}}{\mathit{a}}_{\mathbf{x}}{\mathit{t}}^{\mathbf{2}}$

Here,${\mathit{u}}_{\mathbf{x}}$is the$\mathit{x}$component of the initial velocity,${\mathit{a}}_{\mathbf{x}}$is the acceleration in the$\mathit{x}$direction and$\mathit{t}$is the time.

Find the table of the values of |r→|.

a)

The initial position of the projectile is$(x=0,y=0)$, the velocity of the projectile at$t=0$is$12.0\hat{\mathrm{i}}+49.0\hat{\mathrm{j}}$. The value of the acceleration due to gravity is$9.8\mathrm{m}/{\mathrm{s}}^2$.

In a vertical projectile the acceleration in$x$direction is zero.

The $x$coordinate of the projectile is given by,

$x\left(t\right)=u_{x}t+\frac{1}{2}{a}_x{t}^2$

Substitute 12.0 for $u_x$, data-custom-editor="chemistry" $0$for $a_x$in the above equation.

$\begin{array}{c}x\left(t\right)=12.0t+\frac{1}{2}\left(0\right)t^2\\ =12.0t\end{array}$

Thus, the $0$coordinate of the position vector is 12.0 t.

Write the formula to calculate thecoordinate of the projectile

$y\left(t\right)=u_{\mathrm{y}}t+\frac{1}{2}{a}_{\mathrm{y}}{t}^2$

Here,$u_{\mathrm{y}}$is the$y$component of the initial velocity,$a_{\mathrm{y}}$is the acceleration in the$y$direction and$t$is the time. For the vertical projectile the acceleration in$y$direction is the acceleration due to gravity.

Substitute 49.0 for$u_{\mathrm{y}},-9.8\mathrm{m}/{\mathrm{s}}^2$for$a_{\mathrm{y}}$in the above equation.

$\begin{array}{c}y\left(t\right)=(49.0)t+\frac{1}{2}\left(-9.8\mathrm{m}/{\mathrm{s}}^2\right)t^2\\ =(49.0)t-\left(4.9\mathrm{m}/{\mathrm{s}}^2\right)t^2\end{array}$

Thus, the $y$coordinate of the position vector is $(49.0)t-\left(4.9\mathrm{m}/{\mathrm{s}}^2\right)t^2$.

Write the formula to calculate the magnitude of the position vector

$\left|\overrightarrow{\mathrm{r}}\right|=\sqrt{x^2+y^2}$

Substitute 12.0 t for$x$and$(49.0)t-\left(4.9\mathrm{m}/{\mathrm{s}}^2\right)t^2$for$y$in the above equation.

For$0\le t\le 10\mathrm{s}$, the table of the values of$\left|\overrightarrow{\mathrm{r}}\right|$is shown in the table below.

t(sec)	Vec(r)
0	0
1	45.7
2	82.0
3	109
4	127
5	136
6	138
7	133
8	124
9	117
10	120

Prove that the distance is a maximum when the position vector is perpendicular to the velocity.

b)

The initial position of the projectile is$(x=0,y=0)$, the velocity of the projectile at$t=0$is$12.0\hat{\mathrm{i}}+49.0\hat{\mathrm{j}}$.

The velocity vector tells about the change in the position vector. If the velocity vector at particular point has a component along the position vector and the velocity vector makes an angle less than${90}^{\circ }$with position vector, then the position vector magnitude’s increases and if the angle is greater thanthe magnitude of the position vector starts decreasing.

For the position vector to be maximum the distance from the origin must be momentarily at rest or constant and the only possible situation for this is that the velocity vector makes an angle with the position vector.

The magnitude of maximum displacement.

c)

The initial position of the projectile is$(x=0,y=0)$, the velocity of the projectile at$t=0$is$12.0\hat{\mathrm{i}}+49.0\hat{\mathrm{j}}$.

The expression for the position vector

$\left|\overrightarrow{\mathrm{r}}\right|=\sqrt{{(12.0t)}^2+{\left[\left((49.0)t-\left(4.9\mathrm{m}/{\mathrm{s}}^2\right)t^2\right)\right]}^2}$

Square both the sides of the above equation.

role="math" localid="1663668063540" $|\overrightarrow{\mathrm{r}}{|}^2=(12.0t{)}^2+{\left[\left((49.0)t-\left(4.9\mathrm{m}/{\mathrm{s}}^2\right)t^2\right)\right]}^2$

Differentiate the above expression with respect to$t$to calculate the maxima condition for$r^2$.

$\frac{d}{dt}|\overrightarrow{\mathrm{r}}{|}^2=\frac{d}{dt}\left[{(12.0t)}^2+{\left[\left((49.0)t-\left(4.9\mathrm{m}/{\mathrm{s}}^2\right)t^2\right)\right]}^2\right]$

For the maxima condition to calculate the value the value of$t$for which$r^2$is maximum substitute 0 for$\frac{d}{dt}|\overrightarrow{r}{|}^2.$

$\begin{array}{c}0=\frac{d}{dt}\left[{(12.0t)}^2+{\left[\left((49.0)t-(4.9)t^2\right)\right]}^2\right]\\ =24.0t+\left[\left(98.0t-\left(9.8t^2\right)\right)(49.0-9.8\mathrm{t})\right]\end{array}$

Further solve the above expression for.

$\begin{array}{rcl}96.04{\mathrm{t}}^3-1440.6t^2+4826t& =& 0\\ 96.04{\mathrm{t}}^2-1440.6t+4826& =& 0\end{array}$

Solve further the above quadratic equation, the values of$t$are$5.7\text{s}$and$9.9\text{s}$.

From the table in Part (a) it is evident that after$6\text{s}$the distance starts to decrease therefore, for$t=9.9\text{s}$the position vector does not have maximum magnitude.

Thus, the maximum value of $\overrightarrow{\mathrm{r}}(t{)}^2$is at $t=5.7\text{s}$, therefore the maximum value of $\overrightarrow{\mathrm{r}}\left(t\right)$is also at $t=5.7\text{s}$.

Substitute role="math" localid="1663668478316" $5.7\text{s}$for $t$in equation (1).

role="math" localid="1663668711548" $\begin{array}{c}|\overrightarrow{\mathrm{r}}{|}_{\max }=\sqrt{{(12.0(5.70\mathrm{s}\left)\right)}^2+{\left[\left((49.0)(5.70\mathrm{s})-\left(4.9\mathrm{m}/{\mathrm{s}}^2\right){(5.70\mathrm{s})}^2\right)\right]}^2} \\ = 138.23 \text{m}\\ = 138 \text{m}\end{array}$

The maximum distance is $1.38\text{m}$.

The explanation for the method used in part (c) calculation.

d)

The initial position of the projectile is$(x=0,y=0)$, the velocity of the projectile at$t=0$is$12.0\hat{\mathrm{i}}+49.0\hat{\mathrm{j}}$.

The maximum or minimum value of any function is easily calculated the Maxima and Minima condition.

For the part (c) first calculate the critical points by equating the differential of$r^2$to zero to find the values of$t$for which has the maximum value.

$\frac{d}{dt}{r}^2=0$

The value of$t$where$r^2$is maximum has also the maximum value at the same$t$.

Substitute the maximum value of$r$in the above expression to calculate the maximum magnitude of the position vector.