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

Motion in a Straight Line

Solution:

Velocity when the engine is switched off
$v=19.6 \times 5=98 ms ^{-1}$
$h_{\max }=h_1+h_2 \text { where } h_1=\frac{1}{2} a t^2\, \&\, h_2=\frac{v^2}{2 a} $
$h_{\max }=\frac{1}{2} \times 19.6 \times 5 \times 5+\frac{98 \times 98}{2 \times 9.8}=735 \,m$