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

Answer:

$\theta = 49.56 degree$

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

$y = \frac{1}{2}gt^2$

$78 = \frac{1}{2}(9.81)t^2$

$t = 3.98 s$

so the distance of the car must be

$d = v_r\times t$

$d = 16.67 \times 3.98$

$d = 66.47 m$

Angle of the car with horizontal is given as

$tan\theta = \frac{y}{x}$

$tan\theta = \frac{78}{66.47}$

$\theta = tan^{-1}(1.17)$

$\theta = 49.56 degree$