91影视

When a projectile is projected horizontally from a height yabove the ground with initial velocity $v_{x0}$, it moves under the effect of two independent velocities $v_x$and $v_y$. If the starting point is taken as the origin, and the downward direction is taken as the positive y-axis, the horizontal and vertical components of acceleration will be $a_{\mathrm{x}}=0;a_{\mathrm{y}}=\mathrm{g}$. Thus, the kinematics equations for the projectile motion are as follows:

$\begin{array}{c}{v}_{\mathrm{x}}=v_{\mathrm{x}0}\\ x=v_{\mathrm{x}0}t\end{array}$

$\begin{array}{c}{v}_{\mathrm{y}}=v_{\mathrm{y}0}+\mathrm{g}t\\ y=v_{\mathrm{y}0}t+\frac{1}{2}\mathrm{g}{t}^2\\ v_{\mathrm{y}}^2=v_{\mathrm{y}0}^2+2\mathrm{g}y\end{array}$

Here, xandy are the horizontal and vertical displacements of the projectile traveled in time t.

The vertical displacement of the projectile is $y=235\mathrm{m}$.

The initial horizontal velocity of the projectile is $v_{x0}=69.4\mathrm{m}/\mathrm{s}$.

The initial vertical velocity of the projectile is $v_{y0}=0\mathrm{m}/\mathrm{s}$.

Here, the goods thrown by the plane is your projectile. Let the horizontal displacement of the projectile be$x$ and the time taken by the projectile to reach the ground be t.

Using the kinematics equation for the vertical motion of a projectile, you will get the time as

$\begin{array}{c}y=v_{y0}t+\frac{1}{2}\mathrm{g}{t}^2\\ 235\mathrm{m}=\left(0\mathrm{m}/\mathrm{s}\right)t+\frac{1}{2}\left(9.8\mathrm{m}/{\mathrm{s}}^2\right)t^2\\ t^2=\frac{2脳\left(235\mathrm{m}\right)}{\left(9.8\mathrm{m}/{\mathrm{s}}^2\right)}\\ t^2=47.96{\mathrm{s}}^2\\ t=6.93\mathrm{s}\end{array}$

Using the kinematics equation for the horizontal motion of a projectile, you will get the horizontal distance as

$\begin{array}{c}x=v_{x0}t\\ =\left(69.4\mathrm{m}/\mathrm{s}\right)脳\left(6.93\mathrm{s}\right)\\ =480.94\mathrm{m}\end{array}$.

Thus, the horizontal distance traveled by the goods is 480.94 m.