Question

A relief aeroplane is flying at a constant height of 1960m with
speed 600km/hr above the ground towards a point directly over a
person struggling in flood water (see figure). If the angle of sight, at
which the pilot releases a survival kit if it is to reach the person in
water is tan '(x). Find x. (g = 9.8 m/s?)
hematics-1,2 & Calculus
17​

Answer 1

Answer:

${60}^{\circ}$

Explanation:

Plane is flying at speed = $600* \dfrac{5}{18} = \dfrac {500}{3} m {s}^{-1}$ at the height of 1960m

Time Taken By Kit to reach the ground

$t = \sqrt{{\dfrac{2h}{g}}}$

     $= \sqrt{{\dfrac{2*1960}{9.8}}}$

$t = 20s$

in this time the kit will move horizontally by

$x= ut = \dfrac {500}{3}*{20}= \dfrac{1000}{3} m$

So tha angle of sight

$\tan \theta = \dfrac {x}{h} = \dfrac {1000}{3*1960}$

$= \dfrac {10}{5.88}$

$= 1.7$

$= \sqrt{3}$

therefore,

$\theta={60}^{\circ}$

Answer 2

Answer:

60°

Explanation:

refer the attachment.

hope this helps you out.