## Physics: Principles with Applications (7th Edition)

The packet should be released when the angle below the horizontal is $53.6^{\circ}$
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 $53.6^{\circ}$