Question

Using a telescope, you track a rocket that was launched 13 km away recording the angle theta between the telescope and the ground at half‑second intervals. Estimate the velocity of the rocket of theta(10)=0.205 and theta(10.5)=0.225.

Answer 1

Answer:

$\dfrac{dh}{dt} = 1872 km/h$

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,

$tan \theta = \dfrac{h}{d}$

$tan \theta = \dfrac{h}{13}$

$\dfrac{dh}{dt} = 13 sec^2\theta \dfrac{d\theta }{dt}$

${d\theta }{dt} = \dfrac{\theta(10.5)-\theta(10)}{0.5}$

${d\theta }{dt} = \dfrac{0.225-0.205}{0.5}$

${d\theta }{dt} =0.04$

$\dfrac{dh}{dt} = 13 sec^2(0.205) \times 0.04$

$\dfrac{dh}{dt} = 0.52 km/s$

$\dfrac{dh}{dt} = 1872 km/h$