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# Solve the following :

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 $=42+6=48 \mathrm{~m}$