91影视

Given :

Cannon fires a projectile to the right with speed : $v_0$

Barrel elevated at angle :$u$

Speed of train : $v_{train}$

Acceleration :$a$

The velocity components of the projectile are given by

$$\begin{gathered} {\left(v_i\right)}_{x.p}=v_o\cos 胃+v_{train} \\ {\left(v_i\right)}_{y.p}=v_o\sin 胃 \end{gathered}$$

relative to the ground.

The train's motion has no effect on the $y$component of the velocity. The following kinematic equation describes the y component of velocity at time$t\text{'}$

${\left(v_f\right)}_{y.p}={\left(v_i\right)}_{y.p}-gt\text{'}$

When the projectile reaches its maximum height, ${\left(v_f\right)}_{y.p}=0$

As a result, $t\text{'}=\frac{{\left(v_i\right)}_{y.p}}{g}=\frac{v_o\sin 胃}{g}$gives the time it takes for the projectile to reach maximum height.

$t=2t\text{'}=\frac{2v_o\sin 胃}{g}$

is the horizontal distance travelled by the projectile during this time.

The horizontal distance travelled by the train during this period is given by

role="math" localid="1650778947332" $$\begin{gathered} x_p={\left(v_i\right)}_{x.p}t=(v_o\cos 胃+v_{train})\frac{2v_o\sin 胃}{g} \\ =\boxed{\frac{2{{\mathrm{v}}_{\mathrm{o}}}^2\mathrm{肠辞蝉胃蝉颈苍胃}}{\mathrm{g}}+\frac{2{\mathrm{v}}_{\mathrm{train}}{\mathrm{v}}_{\mathrm{o}}\mathrm{蝉颈苍胃}}{\mathrm{g}}} \end{gathered}$$

Time taken by the projectile :

role="math" localid="1650779047723" $$\begin{gathered} x_T=v_{train}t+\frac{at虏}{2} \\ =\frac{2{\mathrm{v}}_{\mathrm{train}}{\mathrm{v}}_{\mathrm{o}}\mathrm{蝉颈苍胃}}{\mathrm{g}}+\frac{4a{{\mathrm{v}}_{\mathrm{o}}}^2{\sin }^2胃}{2{\mathrm{g}}^2} \\ =\frac{\mathbf{2}{\mathbf{v}}_{\mathbf{train}}{\mathbf{v}}_{\mathbf{o}}\mathbf{蝉颈苍胃}}{\mathbf{g}}\mathbf{+}\frac{\mathbf{2}\mathbf{a}{{\mathbf{v}}_{\mathbf{o}}}^{\mathbf{2}}{\mathbf{sin}}^{\mathbf{2}}\mathbf{胃}}{{\mathbf{g}}^{\mathbf{2}}} \end{gathered}$$

$$\begin{gathered} R=x_p-x_T \\ =\frac{{{\mathrm{v}}_{\mathrm{o}}}^2}{g}\left[\sin 2\mathrm{胃}-\frac{2a}{\mathrm{g}}\left(\frac{1-\cos 2\mathrm{胃}}{2}\right)\right] \\ =\frac{{{\mathrm{v}}_{\mathrm{o}}}^2}{g}\left[\sin 2\mathrm{胃}+\frac{a}{\mathrm{g}}\cos 2\mathrm{胃}-\frac{\mathrm{a}}{\mathrm{g}}\right] \end{gathered}$$

The following condition must be used to obtain the value of 0 for which R is maximum.

$$\begin{gathered} \frac{dR}{d胃}=0 \\ 鈬�\frac{dR}{d胃}=\frac{{{\mathrm{v}}_{\mathrm{o}}}^2}{g}\left[2\cos 2\mathrm{胃}-\frac{2a}{\mathrm{g}}\sin 2\mathrm{胃}\right]=0 \\ 鈬�\frac{a}{\mathrm{g}}\sin 2\mathrm{胃}=\cos 2\mathrm{胃} \\ 鈬�\tan 2\mathrm{胃}=\frac{\mathrm{g}}{\mathrm{a}} \\ 鈬�\mathrm{胃}=\boxed{\frac{\mathbf{1}}{\mathbf{2}}{\mathbf{tan}}^{\mathbf{-}\mathbf{1}}\left(\frac{g}{a}\right)} \end{gathered}$$

This is the situation where the projectile is required to be.