Mechanics Questions

Question Number 209359 by peter frank last updated on 08/Jul/24

Commented by mr W last updated on 08/Jul/24

$${with}\:{a}\:{vertical}\:{velocity}\:{v}''\:{should}\:{be} \\$$$${with}\:{a}\:{velocity}\:{v}''. \\$$$${answer}\:\frac{{v}^{\mathrm{2}} −{g}^{\mathrm{2}} {x}^{\mathrm{2}} }{\mathrm{2}{gv}^{\mathrm{2}} }\:{should}\:{be}\:\:\frac{{v}^{\mathrm{4}} −{g}^{\mathrm{2}} {x}^{\mathrm{2}} }{\mathrm{2}{gv}^{\mathrm{2}} }. \\$$

Commented by peter frank last updated on 08/Jul/24

$$\mathrm{true}\:\mathrm{its}\:\mathrm{typing}\:\mathrm{arror}. \\$$

Answered by Spillover last updated on 08/Jul/24

$${y}={x}\mathrm{tan}\:\theta−\frac{{gx}^{\mathrm{2}} }{\mathrm{2}{u}^{\mathrm{2}} }\mathrm{sec}\:^{\mathrm{2}} \theta \\$$$${u}={v} \\$$$${let}\:\mathrm{tan}\:\theta={X} \\$$$${y}={xX}−\frac{{gx}^{\mathrm{2}} }{\mathrm{2}{v}^{\mathrm{2}} }\left(\mathrm{1}+{X}^{\mathrm{2}} \right)......\left({i}\right) \\$$$$\frac{{dy}}{{dX}}={x}−\frac{{gx}^{\mathrm{2}} }{\mathrm{2}{v}^{\mathrm{2}} }\left(\mathrm{2}{X}\right) \\$$$$\frac{{dy}}{{dX}}={x}−\frac{{gx}^{\mathrm{2}} }{\mathrm{2}{v}^{\mathrm{2}} }\left(\mathrm{2}{X}\right)={x}−\frac{{gx}^{\mathrm{2}} }{{u}^{\mathrm{2}} }{X} \\$$$$\frac{{dy}}{{dX}}=\mathrm{0}\:\:\:\:\:\:\:\:\:\:\:\:\:\mathrm{0}={x}−\frac{{gx}^{\mathrm{2}} }{{v}^{\mathrm{2}} }{X}\:\:\:\:\:\:\:\:\:\:\:{X}=\frac{{v}^{\mathrm{2}} }{{gx}}\:\:\:\:\:\:.....\left({ii}\right) \\$$$${y}={xX}−\frac{{gx}^{\mathrm{2}} }{\mathrm{2}{v}^{\mathrm{2}} }\left(\mathrm{1}+{X}^{\mathrm{2}} \right) \\$$$${y}_{{max}} ={x}\frac{{v}^{\mathrm{2}} }{{gx}}\:−\frac{{gx}^{\mathrm{2}} }{\mathrm{2}{v}^{\mathrm{2}} }\left(\mathrm{1}+\frac{{v}^{\mathrm{2}} }{{gx}}\:\right)=\frac{{v}^{\mathrm{4}} −{g}^{\mathrm{2}} {x}^{\mathrm{2}} }{\mathrm{2}{gv}^{\mathrm{2}} } \\$$$${y}_{{max}} =\frac{{v}^{\mathrm{4}} −{g}^{\mathrm{2}} {x}^{\mathrm{2}} }{\mathrm{2}{gv}^{\mathrm{2}} } \\$$

Commented by Spillover last updated on 08/Jul/24

$${method}\:\mathrm{1} \\$$

Commented by peter frank last updated on 08/Jul/24

$$\mathrm{thanks} \\$$

Answered by Spillover last updated on 08/Jul/24

$${from}\:{equation}\:{of}\:{trajactory} \\$$$${y}={x}\mathrm{tan}\:\theta−\frac{{gx}^{\mathrm{2}} }{\mathrm{2}{v}^{\mathrm{2}} }\mathrm{sec}\:^{\mathrm{2}} \theta\:\:\:\:....\left({i}\right) \\$$$$\frac{{dy}}{{d}\theta}={x}\mathrm{sec}\:^{\mathrm{2}} \theta−\frac{{gx}^{\mathrm{2}} }{\mathrm{2}{v}^{\mathrm{2}} }\mathrm{2sec}\:^{\mathrm{2}} \theta\mathrm{tan}\:\theta \\$$$$\mathrm{0}={x}\mathrm{sec}\:^{\mathrm{2}} \theta−\frac{{gx}^{\mathrm{2}} }{\mathrm{2}{v}^{\mathrm{2}} }\mathrm{2sec}\:^{\mathrm{2}} \theta\mathrm{tan}\:\theta \\$$$$\mathrm{tan}\:\theta=\frac{{v}^{\mathrm{2}} }{{gx}^{\mathrm{2}} }\:\:\:\:.....\left({ii}\right) \\$$$${from}\:\left({i}\right) \\$$$${y}={x}\mathrm{tan}\:\theta−\frac{{gx}^{\mathrm{2}} }{\mathrm{2}{v}^{\mathrm{2}} }\mathrm{sec}\:^{\mathrm{2}} \theta\:\:\:\: \\$$$${y}_{{max}} ={x}\frac{{v}^{\mathrm{2}} }{{gx}^{\mathrm{2}} }\:\:−\frac{{gx}^{\mathrm{2}} }{\mathrm{2}{v}^{\mathrm{2}} }\left(\mathrm{1}+\frac{{v}^{\mathrm{2}} }{{gx}^{\mathrm{2}} }\:\right) \\$$$${y}_{{max}} ={x}.\frac{{v}^{\mathrm{2}} }{{gx}^{\mathrm{2}} }−\frac{{gx}^{\mathrm{2}} }{\mathrm{2}{v}^{\mathrm{2}} }\left(\mathrm{1}+\frac{{v}^{\mathrm{2}} }{{gx}^{\mathrm{2}} }\right) \\$$$${y}_{{max}} ={x}.\frac{{v}^{\mathrm{2}} }{{gx}^{\mathrm{2}} }−\frac{{gx}^{\mathrm{2}} }{\mathrm{2}{v}^{\mathrm{2}} }\left(\mathrm{1}+\frac{{v}^{\mathrm{2}} }{{gx}^{\mathrm{2}} }\right) \\$$$${y}_{{max}} ={x}.\frac{{v}^{\mathrm{2}} }{{gx}^{\mathrm{2}} }−\frac{{gx}^{\mathrm{2}} }{\mathrm{2}{v}^{\mathrm{2}} }−\frac{{gx}^{\mathrm{2}} {v}^{\mathrm{2}} }{\mathrm{2}{v}^{\mathrm{2}} {gx}^{\mathrm{2}} } \\$$$${y}_{{max}} =\frac{\mathrm{2}{v}^{\mathrm{4}} −{g}^{\mathrm{2}} {x}^{\mathrm{2}} −{v}^{\mathrm{4}} }{\mathrm{2}{v}^{\mathrm{2}} {g}} \\$$$${y}_{{max}} =\frac{{v}^{\mathrm{4}} −{g}^{\mathrm{2}} {x}^{\mathrm{2}} }{\mathrm{2}{v}^{\mathrm{2}} {g}} \\$$$$\\$$

Commented by Spillover last updated on 08/Jul/24

$${method}\:\mathrm{2} \\$$

Answered by mr W last updated on 08/Jul/24

Commented by mr W last updated on 08/Jul/24

$${x}={v}\:\mathrm{cos}\:\theta\:{t}\:\Rightarrow{t}=\frac{{x}}{{v}\:\mathrm{cos}\:\theta} \\$$$${h}={v}\:\mathrm{sin}\:\theta\:{t}−\frac{{gt}^{\mathrm{2}} }{\mathrm{2}}={x}\:\mathrm{tan}\:\theta−\frac{{gx}^{\mathrm{2}} \left(\mathrm{1}+\mathrm{tan}^{\mathrm{2}} \:\theta\right)}{\mathrm{2}{v}^{\mathrm{2}} } \\$$$${let}\:\lambda=\mathrm{tan}\:\theta \\$$$${h}={x}\lambda−\frac{{gx}^{\mathrm{2}} \left(\mathrm{1}+\lambda^{\mathrm{2}} \right)}{\mathrm{2}{v}^{\mathrm{2}} } \\$$$$\frac{{dh}}{{d}\lambda}={x}−\frac{{gx}^{\mathrm{2}} \mathrm{2}\lambda}{\mathrm{2}{v}^{\mathrm{2}} }=\mathrm{0}\: \\$$$$\Rightarrow\lambda=\frac{{v}^{\mathrm{2}} }{{gx}},\:{i}.{e}.\:\:\theta=\mathrm{tan}^{−\mathrm{1}} \frac{{v}^{\mathrm{2}} }{{gx}} \\$$$${h}_{{max}} ={x}×\frac{{v}^{\mathrm{2}} }{{gx}}−\frac{{gx}^{\mathrm{2}} }{\mathrm{2}{v}^{\mathrm{2}} }\left(\mathrm{1}+\frac{{v}^{\mathrm{4}} }{{g}^{\mathrm{2}} {x}^{\mathrm{2}} }\right) \\$$$$\Rightarrow{h}_{{max}} =\frac{{v}^{\mathrm{4}} −{g}^{\mathrm{2}} {x}^{\mathrm{2}} }{\mathrm{2}{gv}^{\mathrm{2}} }\:\checkmark \\$$$${with}\:{x}\leqslant\:{maximum}\:{range}\:\frac{{v}^{\mathrm{2}} }{{g}} \\$$

Commented by peter frank last updated on 08/Jul/24

$$\mathrm{appriciate}\:\mathrm{sir} \\$$