A rocket of height $\displaystyle h$ meters is fired vertically upwards.Its velocity at time t seconds is $\displaystyle (2t+3)$ meters.second.If the angle of elevation of the top of the rocket from a point on the ground after 1 second of firing is $\displaystyle \pi /6$ and after 3 seconds it is $\displaystyle \pi /3$ then the distance of the point from the rocket is :

Integrate the velocity function v = 2t + 3 from 0 to 1 to get the height after 1 second, which is 4 meters, plus the initial height h, making the total height h + 4. Integrating from 0 to 3 gives a height of 18 meters, making the total height h + 18. Let the horizontal distance of the point from the rocket be x, so tan(pi/6) = (h + 4)/x and tan(pi/3) = (h + 18)/x. Solving these equations, x = square root of 3 (h + 4) and x = (h + 18)/square root of 3, which yields h = 11, and substituting back gives x = 7 square root of 3 meters.

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