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Q. A child is swinging a swing. Minimum and maximum heights of swing from earths surface are $ 0.75\,m $ and $2\,m $ respectively. The maximum velocity of this swing is

Punjab PMETPunjab PMET 2009Oscillations

Solution:

According to law of conservation of energy
$\frac{1}{2}mv^{2}_{\max }=mg(h_{2}-h_{1})$
$\Rightarrow v_{\max}=\sqrt{2g(h_{2}-h_{1})}$
$ =\sqrt{2 \times 10 \times (2-0.75)}$
$ =\sqrt{2 \times 10 \times 1.25}$
$=\sqrt{25}=5\,ms^{-1}$