Final answer:
The speed of the aircraft is 300 m/s.
Step-by-step explanation:
First, we need to find the time of flight by using the vertical motion equation:
h = ut + (1/2)gt^2
Where h is the altitude, u is the initial vertical velocity, g is the acceleration due to gravity, and t is the time.
Plugging in the given values: h = 730m, u = 0 m/s (as the mail bag is dropped vertically), and g = 9.8 m/s^2:
730 = 0 + (1/2) * 9.8 * t^2
Simplifying the equation, we get:
t^2 = (2 * 730) / 9.8
t^2 = 148.98
t = sqrt(148.98)
t = 12.2s
Next, we can find the horizontal displacement (x) using the horizontal motion equation:
x = vt
Where v is the horizontal velocity and t is the time of flight.
Plugging in the given values: x = v * 12.2
And knowing that the angle of descent is 37° and the vertical velocity (v) can be found using trigonometry:
v = 300 * cos(37°)
Plugging in the value of v:
x = 300 * cos(37°) * 12.2
Finally, we can calculate the speed of the aircraft using the formula:
speed = sqrt(v^2 + u^2)
Substituting the values: speed = sqrt((300*sin(37°))^2 + (300*cos(37°))^2)
Simplifying this equation: speed = sqrt(300^2 * (sin^2(37°) + cos^2(37°)))
Since sin^2(37°) + cos^2(37°) = 1, the equation simplifies to: speed = sqrt(300^2) = 300 m/s