a) Time of flight: 22.6 s
To calculate the time it takes for the cargo to reach the ground, we just consider the vertical motion of the cargo.
The vertical position at time t is given by
![y(t) = h +u_y t - \frac{1}{2}gt^2](https://tex.z-dn.net/?f=y%28t%29%20%3D%20h%20%2Bu_y%20t%20-%20%5Cfrac%7B1%7D%7B2%7Dgt%5E2)
where
h = 2.5 km = 2500 m is the initial height
is the initial vertical velocity of the cargo
g = 9.8 m/s^2 is the acceleration of gravity
The cargo reaches the ground when
![y(t) = 0](https://tex.z-dn.net/?f=y%28t%29%20%3D%200)
So substituting it into the equation and solving for t, we find the time of flight of the cargo:
![0 = h - \frac{1}{2}gt^2\\t=\sqrt{\frac{2h}{g}}=\sqrt{\frac{2(2500)}{9.8}}=22.6 s](https://tex.z-dn.net/?f=0%20%3D%20h%20-%20%5Cfrac%7B1%7D%7B2%7Dgt%5E2%5C%5Ct%3D%5Csqrt%7B%5Cfrac%7B2h%7D%7Bg%7D%7D%3D%5Csqrt%7B%5Cfrac%7B2%282500%29%7D%7B9.8%7D%7D%3D22.6%20s)
b) 7.5 km
The range travelled by the cargo can be calculated by considering its horizontal motion only. In fact, the horizontal motion is a uniform motion, with constant velocity equal to the initial velocity of the jet:
![v_x = 1200 km/h \cdot \frac{1000 m/km}{3600 s/h}=333.3 m/s](https://tex.z-dn.net/?f=v_x%20%3D%201200%20km%2Fh%20%5Ccdot%20%5Cfrac%7B1000%20m%2Fkm%7D%7B3600%20s%2Fh%7D%3D333.3%20m%2Fs)
So the horizontal distance travelled is
![d=v_x t](https://tex.z-dn.net/?f=d%3Dv_x%20t)
And if we substitute the time of flight,
t = 22.6 s
We find the range of the cargo:
![d=(333.3)(22.6)=7533 m = 7.5 km](https://tex.z-dn.net/?f=d%3D%28333.3%29%2822.6%29%3D7533%20m%20%3D%207.5%20km)