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When a projectile is projected horizontally from a height yabove the ground with an initial velocity$v_{x0}$, it falls downward in a parabolic path under the effect of gravity. If the initial point is taken as the origin and downward direction is taken as positive y-axis, then the kinematic equations for projectile motion will be as follows:

$\begin{array}{c}{v}_x=v_{x0}\\ x=v_{x0}t\end{array}$

$\begin{array}{c}{v}_y=v_{y0}+\mathrm{g}t\\ y=v_{y0}t+\frac{1}{2}\mathrm{g}{t}^2\\ v_y^2=v_{y0}^2+2\mathrm{g}y\end{array}$

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

Consider that the supplies which are thrown by the plane are projectiles.

The horizontal distance of the projectile is $x=425\mathrm{m}$.

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

The initial horizontal velocity of the projectile will be the same as that of the plane, i.e., $v_{x0}=69.4\mathrm{m}/\mathrm{s}$.

Let the initial vertical velocity of the projectile be $v_{y0}$and time taken by the projectile to reach the ground be t.

Using the kinematic equation for the horizontal motion of the projectile, you will get time t as:

$\begin{array}{c}x=v_{x0}t\\ t=\frac{x}{v_{x0}}\\ =\frac{\left(425\mathrm{m}\right)}{\left(69.4\mathrm{m}/\mathrm{s}\right)}\\ =6.12\mathrm{s}\end{array}$

Using the kinematic equation for vertical motion of the projectile, you will get:

$\begin{array}{c}y=v_{y0}t+\frac{1}{2}\mathrm{g}{t}^2\\ 235\mathrm{m}=v_{y0}\left(6.12\mathrm{s}\right)+\frac{1}{2}\left(9.8\mathrm{m}/{\mathrm{s}}^2\right){\left(6.12\mathrm{s}\right)}^2\\ v_{y0}=\frac{\left(235\mathrm{m}\right)}{\left(6.12\mathrm{s}\right)}-\frac{1}{2}\frac{\left(9.8\mathrm{m}/{\mathrm{s}}^2\right){\left(6.12\mathrm{s}\right)}^2}{\left(6.12\mathrm{s}\right)}\\ =38.40\mathrm{m}/\mathrm{s}-29.99\mathrm{m}/\mathrm{s}\\ =8.41\mathrm{m}/\mathrm{s}\end{array}$

Thus, the supplies should be dropped by giving an initial vertical velocity of$8.41\mathrm{m}/\mathrm{s}$ in a downward direction.

Using the kinematic equation for the vertical motion of projectile, you will get:

$\begin{array}{c}{v}_y=v_{y0}+\mathrm{g}t\\ =\left(8.41\mathrm{m}/\mathrm{s}\right)+\left(9.8\mathrm{m}/{\mathrm{s}}^2\right)\left(6.12\mathrm{s}\right)\\ =68.39\mathrm{m}/\mathrm{s}\end{array}$

The magnitude of the final velocity of the supplies, i.e., final speed of the supplies, will be:

$\begin{array}{c}v=\sqrt{v_x^2+v_y^2}\\ =\sqrt{{\left(69.4\mathrm{m}/\mathrm{s}\right)}^2+{\left(68.39\mathrm{m}/\mathrm{s}\right)}^2}\\ =\sqrt{9493.55}\\ =97.44\mathrm{m}/\mathrm{s}\end{array}$

Thus, the supplies land on the ground with a speed of $97.44\mathrm{m}/\mathrm{s}$.