Question

# A body is projected vertically downward from a top of tower of h=40m with velocity =5m/s Another body is projected up from the bottom of tower with vertically upward velocity =15m/s .When &where will bodies meet?

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Solution

## $Letthebodiesmeetaftertseconds.\phantom{\rule{0ex}{0ex}}Signconvention:\phantom{\rule{0ex}{0ex}}Up:+veandDown:-ve\phantom{\rule{0ex}{0ex}}Dis\mathrm{tan}cetravelledbyverticallyupwardprojectedbody:\phantom{\rule{0ex}{0ex}}{s}_{1}=ut-\frac{1}{2}g{t}^{2}\phantom{\rule{0ex}{0ex}}=15t-\frac{1}{2}×9.8{t}^{2}\phantom{\rule{0ex}{0ex}}=15t-4.9{t}^{2}\phantom{\rule{0ex}{0ex}}Distancetravelledbyverticallydownwardprojectedbody:\phantom{\rule{0ex}{0ex}}{s}_{2}=ut+\frac{1}{2}g{t}^{2}\phantom{\rule{0ex}{0ex}}=5t+\frac{1}{2}×9.8{t}^{2}\phantom{\rule{0ex}{0ex}}=5t+4.9{t}^{2}\phantom{\rule{0ex}{0ex}}{s}_{1}+{s}_{2}=40\phantom{\rule{0ex}{0ex}}20t=40\phantom{\rule{0ex}{0ex}}\mathbf{t}\mathbf{=}\mathbf{2}\mathbf{}\mathbf{s}\phantom{\rule{0ex}{0ex}}Dis\mathrm{tan}ceofmeetingpointfromtheground:\phantom{\rule{0ex}{0ex}}{s}_{1}=15t-4.9{t}^{2}\phantom{\rule{0ex}{0ex}}=15×2-\left(4.9×4\right)\phantom{\rule{0ex}{0ex}}\mathbf{=}\mathbf{10}\mathbf{.}\mathbf{4}\mathbf{}\mathbf{m}\phantom{\rule{0ex}{0ex}}$

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