Question

If a body of mass $3kg$ is dropped from the top of a tower of height $500m$. Then its kinetic energy after $3s$ is:

• A. $557J$
  
• B. $735J$
  
• C. $1350J$
  
• D. $1296J$
  

Answer 1

The correct option is C $1350J$

The time at which body touches ground is $t=\sqrt{2d/g}=\sqrt{(2\ast 500)/10}=10s$

Therefore $v=gt=10\ast 3=30m/s$
Therefore $KE=\frac{1}{2}\ast m\ast v^2=\frac{1}{2}\ast 3kg\ast {30}^2=1350J$