Rocket goes uupppp!!!!!
A rocket is fired vertically and tracked by the radar R as shown.
At a particular angular position with angle of elevation(theta) from radar = 60., measured parameters are r=9 km, dr/dt = 10m/s and d(theta)/dt = 0.002 rad/s.
The velocity of rocket at this position is:
The answer is 17.7.
This section requires Javascript.
You are seeing this because something didn't load right. We suggest you, (a) try
refreshing the page, (b) enabling javascript if it is disabled on your browser and,
finally, (c)
loading the
non-javascript version of this page
. We're sorry about the hassle.
1 solution
This question is very very simple. I'll use "@" instead of "theta".
Now, y = rSin@ dy/dt = dr/dt.Sin@ + (rCos@)d@/dt = Velocity of rocket
Sunbstituting the given values, we have
velocity of rocket = (10)Sin60 + (9000)(Cos60)(0.002) = 17.7 m/s A better solution is welcomed