Question

A graphing surprise. At time$\mathit{t}\mathbf{=}\mathbf{0}$, a burrito is launched from level ground, with an initial speed of$\mathbf{16}\mathbf{.}\mathbf{0}\mathbf{}\mathbf{m}\mathbf{}\mathbf{/}\mathbf{s}$and launch angle. Imagine a position vector continuously directed from the launching point to the burrito during the flight. Graph the magnitude rof the position vector for (a)${\mathit{\theta }}_{\mathbf{0}}\mathbf{=}\mathbf{40}\mathbf{.}{\mathbf{0}}^{\mathbf{o}}$and (b)${\mathit{\theta }}_{\mathbf{0}}\mathbf{=}\mathbf{80}\mathbf{.}{\mathbf{0}}^{\mathbf{o}}$. For${\mathit{\theta }}_{\mathbf{0}}\mathbf{=}\mathbf{40}\mathbf{.}{\mathbf{0}}^{\mathbf{o}}$, (c) when does rreach its maximum value, (d) what is that value, and how far (e) horizontally and (f) vertically is the burrito from the launch point? For${\mathit{\theta }}_{\mathbf{0}}\mathbf{=}\mathbf{80}\mathbf{.}{\mathbf{0}}^{\mathbf{o}}$(g) when does rreach its maximum value, (h) what is that value, and how far (i) horizontally and (j) vertically is the burrito from the launch point?

Short answer

(a) The graph for magnitude r of the position vector for ${\theta }_0={40}^o$is drawn.

(b) The graph for magnitude r of the position vector for ${\theta }_0={80}^o$is drawn.

(c) For${\theta }_0={40}^o$, r is maximum at 2.1 sec.

(d) Maximum value of r for ${\theta }_0={40}^o$is 25.7 m.

(e) Horizontally the burrito is 25.7 m far from the launch point.

(f) Vertically the burrito is far from the launch pointis 0

(g) For ${\theta }_0={80}^o$, r is maximum at 1.71 sec.

(h) Maximum value of r for ${\theta }_0={80}^o$is 13.5 m.

(i) Horizontally the burrito is. 4.75 m far from the launch point.

(j) Vertically the burrito is 12.6 m far from the launch point.

Step-by-step solution

Given information 

$t=0\mathrm{s}$

$v_0=16\mathrm{m}/\mathrm{s}$

Determining the concept

The problem is based on the standard kinematic equations that describe the motion of an object with constant acceleration.Using the equation for position, the velocity in terms of t by differentiating it can be found. Further, for second particle, using the equation for acceleration, the equation for the velocity for second particle can be found by integrating this. Finally, equating the two equations the time can be computed.

Formulae:

$s=v_{0}t+\frac{1}{2}a{t}^2$

$r=\sqrt{{(x-x_0)}^2+{\left(y-y_0\right)}^2}$

where, sis total displacement, $v_0$is an initial velocity, t is time, role="math" localid="1657095421117" $r,x,x_0,y,y_0$are position vectors and a is an acceleration.

(a) Determining the graph for magnitude r of the position vector for θ0=400

Now,

The displacement in x direction is given by equation (i),

$x=v_{0}t+\frac{1}{2}a{t}^2$

For horizontal motion,

$x-x_0=\left(v_0\cos {\theta }_0\right)t$

For vertical motion,

$y-y_0=\left(v_0\sin {\theta }_0\right)t-\frac{1}{2}g{t}^2$

Squaring and adding both equations of horizontal and vertical motions,

$r=\sqrt{{\left(x-x_0\right)}^2{\left(y-y_0\right)}^2}$

$r=\sqrt{{\left[(v_0\cos \theta )t\right]}^2{\left[(v_0\sin \theta )t-\frac{1}{2}g{t}^2\right]}^2}$

$r=t\sqrt{v_0^2-v_{0}g\sin {\theta }_{0}t-\frac{g^2{t}^2}{4}}$

(b) Determining the graph for magnitude r of the position vector for 

From this equation, magnitude of r for ${\theta }_0={80}^0$can be found.

(c) Determining the time for θ0=400, when r is maximum.

Calculation for time, when r is maximum.

Differentiating with respect to t,

$\frac{dr}{dt}=\frac{v_0^2-\left({\displaystyle \frac{3v_{0}gt\sin {\theta }_0}{2}}\right)+\left({\displaystyle \frac{g^2{t}^2}{2}}\right)}{\sqrt{v_0^2-v_{0}g\sin {\theta }_{0}t+{\displaystyle \frac{g^2{t}^2}{2}}}}$

When r reaches maximum value,$\frac{dr}{dt}=0$,

'$0=\frac{v_0^2-\left({\displaystyle \frac{3v_{0}gt\sin {\theta }_0}{2}}\right)+\left({\displaystyle \frac{g^2{t}^2}{2}}\right)}{\sqrt{v_0^2-\sin {\theta }_{0}t+{\displaystyle \frac{g^2{t}^2}{2}}}}$

$0=v_0^2-\left(\frac{3v_{0}gt\sin {\theta }_0}{2}\right)+\left(\frac{g^2{t}^2}{2}\right)$

It is known that,

localid="1657099737652" $v_0=16\mathrm{m}/\mathrm{s}$and localid="1657099746691" $\theta ={40}^0$

$48t^{{}^2}-15t+256=0$

So,

$t=\frac{2v_0\sin {\theta }_0}{g}$

localid="1657100581198" $\frac{2\left(16\right)(\sin {40}^0)}{9.8}$

$2.10s$

Therefore, for ${\theta }_0={40}^o$, r is maximum at $2.10s$.

(d) Determining the maximum value of r for θ0=400

Calculation for maximum value for r at ${\theta }_0={40}^0$

$r=t\sqrt{v_0^2-v_{0}g\sin {\theta }_{0}t-\frac{g^2{t}^2}{2}}$

$r=2.1\sqrt{{16}^2-16\times 9.81\sin 40\times 2.1-\frac{9.8^2\times 2.1^2}{2}}$

$r=25.7\mathrm{m}$

Therefore, the maximum value for r at is ${\theta }_0={40}^0$is $r=25.7\mathrm{m}$

(e) Determining the how far horizontally is the burrito from the launch point

Calculations for horizontal distance are as follow:

$r_x=v_0\cos {\theta }_0\times t$

$r_x=16\times \cos 40\times 2.1$

$r_x=25.7\mathrm{m}$

Therefore, the horizontal distance is $r_x=25.7\mathrm{m}$

(f) Determining the how far vertically is the burrito from the launch point

From above, vertical distance is,
$r_y=0$

(g) Determining time when r is maximum

Calculation of maximum r at$\theta ={80}^o$

${48}^2-15t+256=0$

Solving this equation,

$t=1.71\sec$

Therefore, time when r is maximum is$t=1.71\mathrm{s}$$t=1.71s$

(h) Determining the maximum value of r

Calculations for distance travelled are as follow:

$r=t\sqrt{v_0^2-v_{0}g\sin {\theta }_{0}t-\frac{g^2{t}^2}{4}}$

$r=1.71\sqrt{{16}^2-16\times 9.81\times \sin 40\times 1.71-\frac{9.{81}^{2}1.{71}^2}{4}}$

$r=13.5\mathrm{m}$

Thus, the maximum value of r is$r=13.5\mathrm{m}$

(i) Determining how far horizontally is the burrito from the launch point

Horizontal distance,

$r_x=v_{0\times }\cos \theta \times t$

$$\begin{gathered} r_x=16\times \cos 40\times 1.71 \\ {r}_x=4.75\mathrm{m} \end{gathered}$$

Therefore, the horizontal distance is $r_x=4.75\mathrm{m}$

(j) Determining how far vertically is the burrito from the launch point

Vertical distance,

$r_y=v_0\sin {\theta }_{0}t-0.5\times g\times t^2$

$r_y=16\times \sin 80\times 1.71-0.5\times 9.81\times 1.{71}^2$

$r_y=12.6\mathrm{m}$

Therefore, the vertical distance is$r_y=12.6\mathrm{m}$$r_y=12.6\mathrm{m}$