question_answer

A ball is dropped from a height. It takes 0.2s to cross the last 6 m before hitting the ground. What is the height from which it dropped? Take \[\text{g}=\text{1}0\text{m}/{{\text{s}}^{\text{2}}}\]

Answer:

From \[s=ut+\frac{1}{2}a{{t}^{2}}\] \[6={{u}_{1}}\times 0.2+\frac{1}{2}\times 10\times {{(0.2)}^{2}}\] \[{{u}_{1}}=\frac{5.8}{0.2}=29m/s\] from \[{{v}^{2}}-{{u}^{2}}=2as,\,\,{{v}^{2}}=u_{1}^{2}+2gs\] \[={{(29)}^{2}}+2\times 10\times 6=841+120=961\,\,m/{{s}^{2}}\] but the height from which the ball was dropped, \[h=\frac{{{V}^{2}}}{2g}=\frac{961}{2\times 10}=48.05\,m\approx 48m\]