Question

The pilot of an airplane carrying a package of mail to a remote outpost wishes to release the package at the right moment to hit the recovery location A. What angle θ with the horizontal should the pilot’s line of sight to the target make at the instant of release? The airplane is flying horizontally at an altitude of 86 m with a velocity of 283 km/h.

Answer 1

The angle is
`\theta\approx 14.61` degrees.

Step-by-step explanation:

Se the attached drawing if you need a visual aid for the explanation. Let
`\theta` be the angle of elevation of the plante which in itself is the same drop angle that the pilot measures. Let
`d` be the horizontal distance from the target and
`h` 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:

`y(t)=-(1)/(2)gt^2+h`

If we set
`y(t)=0` 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.

`0=-(1)/(2)gt^2+h\implies t_f=\sqrt{(2h)/(g)}`.

If the flight time is -
`t_f` then the distance b can be found in meters by taking into account that the horizontal speed of the plane is
`v=283\, Km/h=78.61 \, m/s`.

`d=v\cdot t_f=78.61\cdot \sqrt{(2h)/(g)}`

The angle is thus

`\theta=\arctan{(h)/(v\cdot t_f)}=\arctan{\frac{h}{v\cdot \sqrt{(2\cdot h)/(g)}}\approx 14.61` degrees.