91Ó°ÊÓ

Initial speed of airplane is

$$\begin{gathered} {\mathrm{v}}_0=290\frac{\mathrm{km}}{\mathrm{hr}} \\ =80.6\frac{\mathrm{m}}{\mathrm{s}} \end{gathered}$$

Projection angle is${}^{\mathrm{θ}=-30°}$below the horizontal.

Horizontal distance between release point of the decoy and the point where decoy strike the ground is${}^{d_x=700m}$

This problem deals with kinematic equations that describe the motion of an object with constant acceleration. Using the standard equation for the velocity of the object, the time that the decoy spent in air can be computed. Further, using the formula for the second kinematic equation, the height of release point can be found.

Formula:

The displacement for the horizontal and vertical direction can be written as,

$d_x=\left(v_0\cos \mathrm{θ}\right)t$ (i)

$d_y-d_{y0}=\left(v_0\sin \mathrm{θ}\right)t-\frac{1}{2}g{t}^2$

(ii)

Now, using equation (i) the time will be,

$$\begin{gathered} \mathrm{t}=\frac{{\mathrm{d}}_{\mathrm{x}}}{{\mathrm{v}}_0\mathrm{³¦´Ç²õθ}} \\ \mathrm{t}=\frac{700\mathrm{m}}{\left(80.6\mathrm{m}/\mathrm{s}\right)\cos \left(-30.0°\right)} \\ \mathrm{t}=10.02842\mathrm{s} \\ ≈10.0\mathrm{s} \end{gathered}$$

Now,

$$\begin{gathered} {\mathrm{d}}_{\mathrm{y}}-{\mathrm{d}}_{\mathrm{y}0}=\left({\mathrm{v}}_0\mathrm{²õ¾\pm ²Ôθ}\right)\mathrm{t}-\frac{1}{2}{\mathrm{gt}}^2 \\ \mathrm{As}{\mathrm{d}}_{\mathrm{y}}=0\mathrm{m}, \\ {\mathrm{d}}_{\mathrm{y}}-{\mathrm{d}}_{\mathrm{y}0}=\left(80.6\right)\sin \left(-30.0°\right)\left(10.02842\mathrm{s}\right)-\frac{1}{2}\left(9.8\frac{\mathrm{m}}{{\mathrm{s}}^2}\right){\left(10.02842\mathrm{s}\right)}^2 \\ {\mathrm{d}}_{\mathrm{y}0}=896.934\mathrm{m} \\ ≈897\mathrm{m} \end{gathered}$$