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
[tex]y(t) = h +u_y t - \frac{1}{2}gt^2[/tex]
where
h = 2.5 km = 2500 m is the initial height
[tex]u_y = 0[/tex] 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
[tex]y(t) = 0[/tex]
So substituting it into the equation and solving for t, we find the time of flight of the cargo:
[tex]0 = h - \frac{1}{2}gt^2\\t=\sqrt{\frac{2h}{g}}=\sqrt{\frac{2(2500)}{9.8}}=22.6 s[/tex]
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:
[tex]v_x = 1200 km/h \cdot \frac{1000 m/km}{3600 s/h}=333.3 m/s[/tex]
So the horizontal distance travelled is
[tex]d=v_x t[/tex]
And if we substitute the time of flight,
t = 22.6 s
We find the range of the cargo:
[tex]d=(333.3)(22.6)=7533 m = 7.5 km[/tex]
