Answer
The packet should be released when the angle below the horizontal is $53.6^{\circ}$
Work Step by Step
We can find the time for the packet to drop $78.0~m$
$y = \frac{1}{2}at^2$
$t = \sqrt{\frac{2y}{a}}$
$t = \sqrt{\frac{(2)(78.0 ~m)}{(9.80 ~m/s^2)}}$
$t = 3.99 ~s$
The horizontal velocity of the packet is 208 km/h and the car's horizontal velocity is 156 km/h. Therefore, the packet gains on the car at a rate of 52 km/h. We can convert this speed to units of $m/s$.
$v = (52 ~km/h)(\frac{1000 ~m}{1 ~km})(\frac{1 ~h}{3600 ~s})$
$v = 14.44~m/s$
We can find the additional horizontal distance $d$ the packet moves relative to the car in 3.99 seconds.
$d = v~t$
$d = (14.44~m/s)(3.99~s)$
$d = 57.6 ~m$
The helicopter should be 57.6 meters behind the car horizontally when it drops the packet. We have a right triangle with sides of 57.6 m and 78.0 m. We can find the angle $\theta$ below the horizontal.
$tan(\theta) = \frac{78.0~m}{57.6~m}$
$\theta = arctan(\frac{78.0~m}{57.6~m})$
$\theta = 53.6^{\circ}$
The packet should be released when the angle below the horizontal is $53.6^{\circ}$
