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

`(dh)/(dt) = 1872 km/h`

Step-by-step 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 = (h)/(d)`

`tan \theta = (h)/(13)`

`(dh)/(dt) = 13 sec^2\theta (d\theta )/(dt)`

`{d\theta }{dt} = (\theta(10.5)-\theta(10))/(0.5)`

`{d\theta }{dt} = (0.225-0.205)/(0.5)`

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

`(dh)/(dt) = 13 sec^2(0.205) * 0.04`

`(dh)/(dt) = 0.52 km/s`

`(dh)/(dt) = 1872 km/h`