Question 4.2: An airplane is flying horizontally with a constant speed v0 ......
An airplane is flying horizontally with a constant speed v_{0} = 400 km/h at a constant elevation h = 2 km above the ground, see Fig. 4.10. (a) If the pilot decided to release a package of supplies very close to a truck on the ground, then what is the time of flight of the package? (b) What is the horizontal distance covered by the package in that time (which is the same horizontal distance covered by the plane)?
(a) The initial velocity of the package is the same as the velocity of the plane. Therefore, the initial velocity of the package \vec{v}_{0} is horizontal (i.e., θ_{0} = 0) and has a magnitude of 400 km/h. Since we know the vertical distance that the package falls, then we find its time of flight from Eq. 4.20 as follows:
y=v_{y o}t-{\frac{1}{2}}g t^{2} ⇒ y=(v_{\mathrm{o}}\sin\theta_{\mathrm{o}})t-{\textstyle\frac{1}{2}}g t^{2} (4.20)
y=(v_{o}\sin\theta_{o})\;t-{\frac{1}{2}}g t^{2}
Substituting with y = −2,000 m (we use the negative sign because the package is bellow the origin) and θ_{0} = 0, we get:
-2.000\mathrm{~m}=0-{\textstyle\frac{1}{2}}\,(9.8\,\mathrm{m}/{s}^{2})t^{2}Solving for t and taking the positive root yields:
t={\sqrt{\frac{2,000{\mathrm{\ m}}}{4.9\,\mathrm{m}/s^{2}}}}=20.2 \ s(b) The horizontal distance covered by the package in that time is:
x=(v_{o}\cos{\theta_{o}})\,t.= (400 km/h)(1 h/3600 s)(cos 0°)(20.2 s) = 2.244 km
= 2,244 m