Question

A rocket is fired upward from the earth surface such that it creates an acceleration of $19.6m/s^2$. If after $5s$, its engine is switched off, the maximum height of the rocket from earth's surface would be

• A. 245 m
• B. 490 m
• C. 980m
• D. 735m

Answer 1

Answer: (A)

Speed of rocket after $5s$.
$v=u-gt$
$0=u-9.8\times 5$
$=49m/s$
From $h=ut-\frac{1}{2}g{t}^2$
$=0-\frac{1}{2}\times 9.8\times (5{)}^2$
$=\frac{245}{2}m$
When engine is turned off
$v^2=u^2-2gh$
$0=u^2-2gh$
$h=\frac{u^2}{2g}=\frac{49\times 49}{2\times 9.8}=\frac{245}{2}m$
Maximum height from earth surface
$=h_1+h_2=\frac{245}{2}+\frac{245}{2}=245m$