Question

Spymaster Paul, flying a constant 215km/h horizontally in a low-flying helicopter, want to drop secret documents into his contact’s open car which is traveling 155km/h on a level highway 78.0m below. At what angle (to the horizontal) should the car be in his sights when the package is released?

Answer 1

`\theta = 49.56 degree`

Step-by-step explanation:

Relative horizontal velocity of plane with respect to the car is given as

`v_r = v_p - v_c`

so we have

`v_p = 215 km/h`

`v_c = 155 km/h`

now we have

`v_r = 215 - 155`

`v_r = 60 km/h`

`v_r = 16.67 m/s`

Now time taken by the object to drop the vertical height is given as

`y = (1)/(2)gt^2`

`78 = (1)/(2)(9.81)t^2`

`t = 3.98 s`

so the distance of the car must be

`d = v_r* t`

`d = 16.67 * 3.98`

`d = 66.47 m`

Angle of the car with horizontal is given as

`tan\theta = (y)/(x)`

`tan\theta = (78)/(66.47)`

`\theta = tan^(-1)(1.17)`

`\theta = 49.56 degree`