Answer:
[tex]\dfrac{dh}{dt} = 1872 km/h[/tex]
Explanation:
given,
distance of the launch from recording = 13 km
time = 0.5 s
theta(10)=0.205 and theta(10.5)=0.225
take height be 'h'
now,
[tex]tan \theta = \dfrac{h}{d}[/tex]
[tex]tan \theta = \dfrac{h}{13}[/tex]
[tex]\dfrac{dh}{dt} = 13 sec^2\theta \dfrac{d\theta }{dt}[/tex]
[tex]{d\theta }{dt} = \dfrac{\theta(10.5)-\theta(10)}{0.5}[/tex]
[tex]{d\theta }{dt} = \dfrac{0.225-0.205}{0.5}[/tex]
[tex]{d\theta }{dt} =0.04[/tex]
[tex]\dfrac{dh}{dt} = 13 sec^2(0.205) \times 0.04[/tex]
[tex]\dfrac{dh}{dt} = 0.52 km/s[/tex]
[tex]\dfrac{dh}{dt} = 1872 km/h[/tex]
