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 - (1)/(2)gt^2](https://img.qammunity.org/2020/formulas/physics/high-school/l6setfhcbgonc0e28a2wflzj3k9tz5arov.png)
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://img.qammunity.org/2020/formulas/physics/high-school/m5ow9i2y9fyf48952bnza84t0ijkri0dux.png)
So substituting it into the equation and solving for t, we find the time of flight of the cargo:
![0 = h - (1)/(2)gt^2\\t=\sqrt{(2h)/(g)}=\sqrt{(2(2500))/(9.8)}=22.6 s](https://img.qammunity.org/2020/formulas/physics/high-school/72yytlc5676iu0ei93mzyuohrtyyxpj05l.png)
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 (1000 m/km)/(3600 s/h)=333.3 m/s](https://img.qammunity.org/2020/formulas/physics/high-school/a6dj6m7vzvj26br0t6z2xi00tnoqhubokl.png)
So the horizontal distance travelled is
![d=v_x t](https://img.qammunity.org/2020/formulas/physics/high-school/cqfjq3nc3ud20ipbnvt98b0fdqn6u2bs8p.png)
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://img.qammunity.org/2020/formulas/physics/high-school/fhf984piv88wbipfwnv2swjddj3ra59oju.png)