Q. A rocket is fired upward from the earth’s surface such that it creates an acceleration of $19.6\ m/\sec^{2}.$ If after $5\sec\ $its engine is switched off, the maximum height of the rocket from earth’s surface would be
A) $245\ m$
B) $490\ m$
C) $980\ m$
D) $735\ m$
Explanation
(d) Given $a = 19.6m/s^{2} = 2g$
Resultant velocity of the rocket after $5\ sec$
$v = 2g \times 5 = 10g\ m/s$
Height achieved after $5\ sec,h_{1} = \frac{1}{2} \times 2g \times 25 = 245m$
On switching off the engine it goes up to height $h_{2}$ where its velocity becomes zero
$0 = (10g)^{2} - 2gh_{2} \Rightarrow h_{2} = 490m$
$\therefore Total\ height\ of\ rocket = 245 + 490 = 735\ m$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.