Answer:
The angle is [tex]\theta\approx 14.61[/tex] degrees.
Explanation:
Se the attached drawing if you need a visual aid for the explanation. Let [tex]\theta[/tex] be the angle of elevation of the plante which in itself is the same drop angle that the pilot measures. Let [tex]d[/tex] be the horizontal distance from the target and [tex]h[/tex] the height of the plane. We know that the package is dropped without any initial vertical speed, that means that it has a y-position equation of the form:
[tex]y(t)=-\frac{1}{2}gt^2+h[/tex]
If we set [tex]y(t)=0[/tex] we are setting the condition that the package is in the ground. We can then solve for t and get the flight time of the package.
[tex]0=-\frac{1}{2}gt^2+h\implies t_f=\sqrt{\frac{2h}{g}}[/tex].
If the flight time is -[tex]t_f[/tex] then the distance b can be found in meters by taking into account that the horizontal speed of the plane is [tex]v=283\, Km/h=78.61 \, m/s[/tex].
[tex]d=v\cdot t_f=78.61\cdot \sqrt{\frac{2h}{g}}[/tex]
The angle is thus
[tex]\theta=\arctan{\frac{h}{v\cdot t_f}}=\arctan{\frac{h}{v\cdot \sqrt{\frac{2\cdot h}{g}}}\approx 14.61 [/tex] degrees.
