Answer
$t =10.0\mathrm{s}.$
Work Step by Step
Define a coordinate system. Let the origin be at the ground directly below the plane at t=0.
Let up be +y, and horizontally, the direction towards the point where the ground is struck to be +x.
If $\vec{v}_{0}$ is expressed as a magnitude (the speed $v_{0}$) and an angle $\theta_{0}$ (measured from the horizontal),
the particle's equations of motion along the horizontal $x$ axis and vertical $y$ axis are
$\left[\begin{array}{ll}
x-x_{0}=(v_{0}\cos\theta_{0})t & (4- 21)\\
y-y_{0}=(v_{0}\sin\theta_{0})t-\frac{1}{2}gt^{2} & (4- 22)\\
v_{y}=v_{0}\sin\theta_{0}-gt & (4- 23)\\
v_{y}^{2}=(v_{0}\sin\theta_{0})^{2}-2g(y-y_{0}) & (4- 24)
\end{array}\right]$
The angle is below the horizontal, $\theta_{0}=-30.0^{o}$
$v_{0}=290 \displaystyle \frac{km}{h}\times\frac{1000m}{1km}\times\frac{1h}{3600s}=80.6$ m/s
(a)
Solve Eq. 4-$21$ to solve for $t$:
$\Delta x=(v_{0}\cos\theta_{0})t $
$t=\displaystyle \frac{700\mathrm{m}}{(80.6m/s)\cos(-30.0^{\mathrm{o}})}=10.0\mathrm{s}.$
