Question

A rocket of height $h$ meters is fired vertically upwards.Its velocity at time t seconds is $(2t+3)$ 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 $\pi /6$ and after 3 seconds it is $\pi /3$ then the distance of the point from the rocket is :

• A. $14\sqrt{3}$ meters
• B. $7\sqrt{3}$ meters
• C. $2\sqrt{3}$ meters
• D. cannot be found without the value of h

Answer 1

Answer: (B)

$7\sqrt{3}$ meters

$\frac{dx}{dt}=2t+3,\Rightarrow \int dx=\int (2t+3)dt\Rightarrow x=t^2+3t+c$
$\therefore BC=1^2+3\times 1+c=4+cBD=3^2+3\times 3+c=18+c$
$\therefore \frac{h+4+c}{AP}=\tan 30,\frac{h+18+c}{AP}=\tan 60$
$\therefore \frac{14}{AP}=\tan 60-\tan 30=\sqrt{3}-\frac{1}{\sqrt{3}}=\frac{2}{\sqrt{3}}$
$\therefore AP=7\sqrt{3}$