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Question Number 135618 by physicstutes last updated on 14/Mar/21
At what speed should a basket ball player project a ball from a height of 2.1 m above the ground at an angle of 38° to the horizontal so that it just passes through a basket ball net which is at a horizontal distance of 5 m from the initial position of the ball and 3 m above the playground. at determine the balls maximum height above the ground.
$$\:\mathrm{At}\:\mathrm{what}\:\mathrm{speed}\:\mathrm{should}\:\mathrm{a}\:\mathrm{basket}\:\mathrm{ball}\:\mathrm{player}\:\mathrm{project}\:\mathrm{a}\:\mathrm{ball}\:\mathrm{from}\:\mathrm{a}\:\mathrm{height} \\ $$$$\mathrm{of}\:\mathrm{2}.\mathrm{1}\:\mathrm{m} \\ $$$$\mathrm{above}\:\mathrm{the}\:\mathrm{ground}\:\mathrm{at}\:\mathrm{an}\:\mathrm{angle}\:\mathrm{of}\:\mathrm{38}° \\ $$$$\mathrm{to}\:\mathrm{the}\:\mathrm{horizontal}\:\mathrm{so}\:\mathrm{that}\:\mathrm{it}\:\mathrm{just}\:\mathrm{passes}\:\mathrm{through}\:\mathrm{a}\:\mathrm{basket}\:\mathrm{ball}\:\mathrm{net}\:\mathrm{which} \\ $$$$\mathrm{is}\:\mathrm{at}\:\mathrm{a}\:\mathrm{horizontal}\:\mathrm{distance}\:\mathrm{of}\:\mathrm{5}\:\mathrm{m}\:\mathrm{from}\:\mathrm{the}\:\mathrm{initial} \\ $$$$\mathrm{position}\:\mathrm{of}\:\mathrm{the}\:\mathrm{ball}\:\mathrm{and}\:\mathrm{3}\:\mathrm{m}\:\mathrm{above}\:\mathrm{the}\:\mathrm{playground}.\:\mathrm{at}\:\mathrm{determine}\:\mathrm{the}\:\mathrm{balls} \\ $$$$\mathrm{maximum}\:\mathrm{height}\:\mathrm{above}\:\mathrm{the}\:\mathrm{ground}. \\ $$
Answered by mr W last updated on 14/Mar/21
L=5 m h=3−2.1=0.9 m θ=38° (h/L)=tan θ−((gL)/(2u^2 ))(1+tan^2 θ) ⇒u=(√((gL(1+tan^2 θ))/(2(tan θ−(h/L)))))≈8.18 m/s h_(max) =((u^2 sin^2 θ)/(2g))=((L(1+tan^2 θ)sin^2 θ)/(4(tan θ−(h/L))))≈1.27 m H_(max) =2.1+1.27=3.37 m above ground
$${L}=\mathrm{5}\:{m} \\ $$$${h}=\mathrm{3}−\mathrm{2}.\mathrm{1}=\mathrm{0}.\mathrm{9}\:{m} \\ $$$$\theta=\mathrm{38}° \\ $$$$\frac{{h}}{{L}}=\mathrm{tan}\:\theta−\frac{{gL}}{\mathrm{2}{u}^{\mathrm{2}} }\left(\mathrm{1}+\mathrm{tan}^{\mathrm{2}} \:\theta\right) \\ $$$$\Rightarrow{u}=\sqrt{\frac{{gL}\left(\mathrm{1}+\mathrm{tan}^{\mathrm{2}} \:\theta\right)}{\mathrm{2}\left(\mathrm{tan}\:\theta−\frac{{h}}{{L}}\right)}}\approx\mathrm{8}.\mathrm{18}\:{m}/{s} \\ $$$${h}_{{max}} =\frac{{u}^{\mathrm{2}} \mathrm{sin}^{\mathrm{2}} \:\theta}{\mathrm{2}{g}}=\frac{{L}\left(\mathrm{1}+\mathrm{tan}^{\mathrm{2}} \:\theta\right)\mathrm{sin}^{\mathrm{2}} \:\theta}{\mathrm{4}\left(\mathrm{tan}\:\theta−\frac{{h}}{{L}}\right)}\approx\mathrm{1}.\mathrm{27}\:{m} \\ $$$${H}_{{max}} =\mathrm{2}.\mathrm{1}+\mathrm{1}.\mathrm{27}=\mathrm{3}.\mathrm{37}\:{m}\:{above}\:{ground} \\ $$
Commented by physicstutes last updated on 14/Mar/21
perfect sir
$$\mathrm{perfect}\:\mathrm{sir} \\ $$
Commented by mr W last updated on 14/Mar/21

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