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Question-135308




Question Number 135308 by bobhans last updated on 12/Mar/21
Commented by mr W last updated on 12/Mar/21
at t=3 s, the height is 71.65m  at t=7.8009s, the ball hits the ground  max. height is 75.4235m  in 6.1554−1.5997=4.5557s the ball  is above 50m
$${at}\:{t}=\mathrm{3}\:{s},\:{the}\:{height}\:{is}\:\mathrm{71}.\mathrm{65}{m} \\ $$$${at}\:{t}=\mathrm{7}.\mathrm{8009}{s},\:{the}\:{ball}\:{hits}\:{the}\:{ground} \\ $$$${max}.\:{height}\:{is}\:\mathrm{75}.\mathrm{4235}{m} \\ $$$${in}\:\mathrm{6}.\mathrm{1554}−\mathrm{1}.\mathrm{5997}=\mathrm{4}.\mathrm{5557}{s}\:{the}\:{ball} \\ $$$${is}\:{above}\:\mathrm{50}{m} \\ $$
Commented by mr W last updated on 12/Mar/21
Answered by Dwaipayan Shikari last updated on 12/Mar/21
h(t)=((−g)/2)t^2 +38t+(7/4)⇒((dh(t))/dt)∣_(t=0) =38(m/s)  =v_0   h(3)=−((9g)/2)+((463)/4)∼71.657 m  (height at 3 s)  At max height  v=0  So h_(max) =(v_0 ^2 /(2g))=((38^2 )/(2.g))=((722)/g) m∼73.67 m  When it hits the ground   ((−g)/2)t^2 +38t+(7/4)=0⇒t=((76+(√(5776+14g)))/(2g))s∼7.80088 s  When height is 50 m then the time is   −(g/2)t^2 +38t+(7/4)=50⇒gt^2 −76t+((193)/2)=0⇒t=((76+(√(5776−386g)))/(2g))s  ∼4.5557 s
$${h}\left({t}\right)=\frac{−{g}}{\mathrm{2}}{t}^{\mathrm{2}} +\mathrm{38}{t}+\frac{\mathrm{7}}{\mathrm{4}}\Rightarrow\frac{{dh}\left({t}\right)}{{dt}}\mid_{{t}=\mathrm{0}} =\mathrm{38}\frac{{m}}{{s}}\:\:={v}_{\mathrm{0}} \\ $$$${h}\left(\mathrm{3}\right)=−\frac{\mathrm{9}{g}}{\mathrm{2}}+\frac{\mathrm{463}}{\mathrm{4}}\sim\mathrm{71}.\mathrm{657}\:{m}\:\:\left({height}\:{at}\:\mathrm{3}\:{s}\right) \\ $$$${At}\:{max}\:{height}\:\:{v}=\mathrm{0} \\ $$$${So}\:{h}_{{max}} =\frac{{v}_{\mathrm{0}} ^{\mathrm{2}} }{\mathrm{2}{g}}=\frac{\mathrm{38}^{\mathrm{2}} }{\mathrm{2}.{g}}=\frac{\mathrm{722}}{{g}}\:{m}\sim\mathrm{73}.\mathrm{67}\:{m} \\ $$$${When}\:{it}\:{hits}\:{the}\:{ground}\: \\ $$$$\frac{−{g}}{\mathrm{2}}{t}^{\mathrm{2}} +\mathrm{38}{t}+\frac{\mathrm{7}}{\mathrm{4}}=\mathrm{0}\Rightarrow{t}=\frac{\mathrm{76}+\sqrt{\mathrm{5776}+\mathrm{14}{g}}}{\mathrm{2}{g}}{s}\sim\mathrm{7}.\mathrm{80088}\:{s} \\ $$$${When}\:{height}\:{is}\:\mathrm{50}\:{m}\:{then}\:{the}\:{time}\:{is}\: \\ $$$$−\frac{{g}}{\mathrm{2}}{t}^{\mathrm{2}} +\mathrm{38}{t}+\frac{\mathrm{7}}{\mathrm{4}}=\mathrm{50}\Rightarrow{gt}^{\mathrm{2}} −\mathrm{76}{t}+\frac{\mathrm{193}}{\mathrm{2}}=\mathrm{0}\Rightarrow{t}=\frac{\mathrm{76}+\sqrt{\mathrm{5776}−\mathrm{386}{g}}}{\mathrm{2}{g}}{s} \\ $$$$\sim\mathrm{4}.\mathrm{5557}\:{s} \\ $$

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