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?

$t=0\mathrm{s}$

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

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.

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,

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

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

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

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

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 $2.10s$.

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 $r=25.7\mathrm{m}$

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 $r_x=25.7\mathrm{m}$

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

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$

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$r=13.5\mathrm{m}$

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 $r_x=4.75\mathrm{m}$

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}$