Question

While launching a rocket of mass $2\times {10}^4$ kg, the $5\times {10}^{5}N$ is applied for $20s$. Calculate the velocity attained by the rocket at the end of $20s$

• A. $300m{s}^{-1}$
• B. $320m{s}^{-1}$
• C. $500m{s}^{-1}$
• D. $100m{s}^{-1}$

Answer 1

Answer: (C)

$500m{s}^{-1}$

Given,

$u=0m/s$

$m=2\times {10}^{4}kg$

$F=5\times {10}^{5}N$

$t=20s$

The force acting on the rpcket is,

$F=ma$

$a=\frac{F}{m}$

$a=\frac{5\times {10}^5}{2\times {10}^4}=25m/s$

From 1st equation of motion,

$v=u+at$

$v=0+25\times 20$

$v=500m/s$

The correct option is C.