Question:
A ball is dropped from a height. If it takes $0.200$ s to cross the last $6.00 \mathrm{~m}$ before hitting the ground, find the height from which it was dropped. Take $g=10 \mathrm{~m} / \mathrm{s}^{2}$
Solution:
For last $6 \mathrm{~m}$,
$\mathrm{t}=0.2 \mathrm{sec} ; \mathrm{s}=6 \mathrm{~m} ; \mathrm{a}=\mathrm{g}$
$s=u t+\frac{1}{2} a t^{2}$
$6=\mathrm{u}(0.2)+\frac{\frac{1}{2}}{}(\mathrm{~g})(0.2)^{2}$
$\mathrm{u}=29 \mathrm{~m} / \mathrm{s}$
Before last $6 \mathrm{~m}$,
$v^{2}=u^{2}+2 a s$
$(29)^{2}=0^{2}+2(g) s$
$\mathrm{S}_{1}=42 \mathrm{~m}$
Total distance $=42+6=48 \mathrm{~m}$
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