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Q.
If a body of mass $ 3\,\,kg $ is dropped from the top of a tower of height $ 25\,\,m $ , then kinetic energy after $ 3\sec $ will be:
Punjab PMETPunjab PMET 2003Work, Energy and Power
Solution:
Here $:$ mass $m=3\, kg$, height $h=25\, m$ and time $t=3 \,sec$
Final velocity of body after $3\, sec$,
$v=u+g t=0+9.8 \times 3=29.4\, m / s$
Hence, kinetic energy of the body is given by
$=\frac{1}{2} m v^{2}=\frac{1}{2} \times 3 \times(29.4)^{2} $
$\approx 1297 \,J$