Question 389724
If {{{y}}} is the height and {{{a }}} the angle of elevation, then:

{{{dy/dt = 300 km/hr}}} ...(1)

{{{y = 5 *tan(a)}}}....(2)


Differentiating (2) with respect to time:

{{{dy/dt = 5* sec^2(a)(da/dt)}}}


Substituting for {{{dy/dt}}} from (1):


{{{300 = 5 *sec^2(a) (da/dt)}}}


{{{da/dt = (300 / 5 sec^2)a}}}


= {{{60 cos^2(a)}}} ...(2)


When {{{y = 2km}}}:

{{{tan(a) = 2/5}}}

{{{cos^2(a)}}}

= {{{1 / sec^2(a)}}}

= {{{1 / (1 + (2/5)^2))}}}

= {{{1 / ( 1 + (4/25))}}}

= {{{25 / 29}}}

Substituting this value in (2):

{{{da/dt = 60 * (25 / 29)}}}

= {{{(1500 / 29 )rad/hr}}}

= {{{(1500 / (29 * 3600)) (rad/sec)}}}

= {{{5 / (29 * 12)}}}

= {{{(5 / 348) (rad/sec)}}}