Question

A rocket is fired upward from the earth surface such that it creates an acceleration of $19.6\, m / s ^{2}$. If after $5\, s$, 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 $5\, s$.
$v =u-g t$
$0 =u-9.8 \times 5$
$=49\, m / s$
From $h =u t-\frac{1}{2} g t^{2}$
$=0-\frac{1}{2} \times 9.8 \times(5)^{2}$
$=\frac{245}{2}\, m$
When engine is turned off
$v^{2}=u^{2}-2 g h$
$0=u^{2}-2 g h$
$h=\frac{u^{2}}{2 g}=\frac{49 \times 49}{2 \times 9.8}=\frac{245}{2}\, m$
Maximum height from earth surface
$=h_{1}+h_{2}=\frac{245}{2}+\frac{245}{2}=245\, m$