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Question

# Q4: 4. An open lift is moving upward with constant velocity 10 ms. A ball is thrown upward with velocity 20 m/s relative to the ground. Find after what time the ball hits the lift again. (A) 2 s (B) 3 s (C) 4 s (D) 5 s

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Solution

## Dear student, $\mathrm{The}\mathrm{initial}\mathrm{speed}\mathrm{of}\mathrm{the}\mathrm{ball},{\mathrm{u}}_{\mathrm{b}}=20\mathrm{m}/\mathrm{s}\phantom{\rule{0ex}{0ex}}\mathrm{As}\mathrm{the}\mathrm{ball}\mathrm{is}\mathrm{moving}\mathrm{upward}\mathrm{and}\mathrm{the}\mathrm{acceleration}\mathrm{due}\mathrm{to}\mathrm{gravity}\mathrm{is}\mathrm{acting}\mathrm{downward},\mathrm{s}\phantom{\rule{0ex}{0ex}}\mathrm{Acceleration}\mathrm{of}\mathrm{the}\mathrm{ball},{\mathrm{a}}_{\mathrm{b}}=-10\mathrm{m}/{\mathrm{s}}^{2}\phantom{\rule{0ex}{0ex}}\mathrm{Let}\mathrm{in}\mathrm{time}\mathrm{t},\mathrm{the}\mathrm{ball}\mathrm{meets}\mathrm{the}\mathrm{lift}.\phantom{\rule{0ex}{0ex}}\mathrm{The}\mathrm{distance}\mathrm{travelled}\mathrm{by}\mathrm{the}\mathrm{ball}\mathrm{in}\mathrm{time}\mathrm{t},\phantom{\rule{0ex}{0ex}}\mathrm{s}={\mathrm{u}}_{\mathrm{b}}\mathrm{t}+\frac{1}{2}{\mathrm{a}}_{\mathrm{b}}{\mathrm{t}}^{2}\phantom{\rule{0ex}{0ex}}⇒\mathrm{s}=20\mathrm{t}-5{\mathrm{t}}^{2}...\left(1\right)\phantom{\rule{0ex}{0ex}}\mathrm{The}\mathrm{speed}\mathrm{of}\mathrm{the}\mathrm{lift},{\mathrm{u}}_{\mathrm{l}}=10\mathrm{m}/\mathrm{s}\phantom{\rule{0ex}{0ex}}\mathrm{Acceleration}\mathrm{of}\mathrm{the}\mathrm{lift},{\mathrm{a}}_{\mathrm{l}}=0m/s\phantom{\rule{0ex}{0ex}}\mathrm{The}\mathrm{distance}\mathrm{travelled}\mathrm{by}\mathrm{the}\mathrm{l}\phantom{\rule{0ex}{0ex}}\mathrm{s}={\mathrm{u}}_{\mathrm{l}}\mathrm{t}+\frac{1}{2}{\mathrm{a}}_{\mathrm{l}}{\mathrm{t}}^{2}\phantom{\rule{0ex}{0ex}}⇒\mathrm{s}=10\mathrm{t}...\left(2\right)\phantom{\rule{0ex}{0ex}}\mathrm{Subtracting}\mathrm{equation}\left(1\right)\mathrm{from}\mathrm{equation}\left(2\right),\mathrm{we}\mathrm{get}\phantom{\rule{0ex}{0ex}}0=-10\mathrm{t}+5{\mathrm{t}}^{2}\phantom{\rule{0ex}{0ex}}⇒5\mathrm{t}\left(\mathrm{t}-2\right)=0\phantom{\rule{0ex}{0ex}}\mathrm{Therefore},\mathrm{t}=2\mathrm{s}\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\mathrm{Regards}$

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