Question

a stomp rocket is launched straight into the air and is being watched by students 500 meters away. after 2 seconds, when the angle of elevation is the angle is increasing at a rate of 0.4 radians per second. how fast is the stomp rocket rising at that moment?

Answer 1

Using the concepts of pressure, we find that 625m$s^{-1}$ is the speed of the stomp rocket rising at that moment when it is watched by students 500 meters away. after 2 seconds, when the angle of elevation is the angle is increasing at a rate of 0.4 radians per second.

We know that when an object is very far away, at the centre of that object there is an angle enclosed by the observer and that is given

angle=(length of arc)/radius of the arc.

θ=l/r

So, we are given that length of arc is 500 metres and angle is 0.4 radians/sec.

So,=>0.4=500/r

      =>r=500/0.4

       =>r=1250m

Now we are given the time when it is observed i.e. is 2 seconds.

So, we apply distance speed formula.,

which is Distance=Speed × Time

            =>1250=Speed×2

              =>Speed=1250/2

              =>Speed =625m$s^{-1}$.

Hence, when a stomp rocket is launched straight into the air and is being watched by students 500 meters away. after 2 seconds, and the angle of elevation is the angle is increasing at a rate of 0.4 radians per second, the speed of stomp rocket is 625m$s^{-1}$.

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