Question-213890

Question Number 213890 by Tawa11 last updated on 20/Nov/24

Answered by mr W last updated on 21/Nov/24

h=u sin θ t−((gt^2 )/2) t=((u sin θ)/g)(1±(√(1−((2gh)/(u^2 sin^2 θ))))) =((45 sin 52°)/(9.81))(1±(√(1−((2×9.81×12)/(45^2 sin^2 52°))))) =6.87 s / 0.36 s R=u cos θ t =45×cos 52°×6.87 / 0.36 =190.4m / 9.86 m

$${h}={u}\:\mathrm{sin}\:\theta\:{t}−\frac{{gt}^{\mathrm{2}} }{\mathrm{2}} \\ $$$${t}=\frac{{u}\:\mathrm{sin}\:\theta}{{g}}\left(\mathrm{1}\pm\sqrt{\mathrm{1}−\frac{\mathrm{2}{gh}}{{u}^{\mathrm{2}} \:\mathrm{sin}^{\mathrm{2}} \:\theta}}\right) \\ $$$$\:\:=\frac{\mathrm{45}\:\mathrm{sin}\:\mathrm{52}°}{\mathrm{9}.\mathrm{81}}\left(\mathrm{1}\pm\sqrt{\mathrm{1}−\frac{\mathrm{2}×\mathrm{9}.\mathrm{81}×\mathrm{12}}{\mathrm{45}^{\mathrm{2}} \:\mathrm{sin}^{\mathrm{2}} \:\mathrm{52}°}}\right) \\ $$$$\:\:=\mathrm{6}.\mathrm{87}\:{s}\:\:\:\:/\:\mathrm{0}.\mathrm{36}\:{s} \\ $$$${R}={u}\:\mathrm{cos}\:\theta\:{t} \\ $$$$\:\:\:\:=\mathrm{45}×\mathrm{cos}\:\mathrm{52}°×\mathrm{6}.\mathrm{87}\:\:\:\:\:/\:\mathrm{0}.\mathrm{36} \\ $$$$\:\:\:\:=\mathrm{190}.\mathrm{4}{m}\:\:\:\:\:\:\:/\:\mathrm{9}.\mathrm{86}\:{m} \\ $$

Commented by mr W last updated on 21/Nov/24

Commented by Tawa11 last updated on 21/Nov/24

$$\mathrm{Thanks}\:\mathrm{sir}.\:\mathrm{I}\:\mathrm{appreciate}. \\ $$

Commented by mr W last updated on 21/Nov/24

$${what}\:{is}\:{right}? \\ $$

Answered by A5T last updated on 21/Nov/24

u_x =ucosθ; u_y =usinθ;u=45ms^(−1) ; θ=52° H=maximum height;R=range t=time from H to 12m above launching point; T=flight time H−12=((gt^2 )/2)⇒((g(((u^2 sin^2 θ)/g^2 )))/2)−12=((gt^2 )/2) ⇒((u^2 sin^2 θ−24g)/(2g))=((gt^2 )/2)⇒t=((√(u^2 sin^2 θ−24g))/g) ⇒T=((√(u^2 sin^2 θ−24g))/g)+((usinθ)/g) R=u_x T=((ucosθ(√(u^2 sin^2 θ−24g)))/g)+((u^2 sin2θ)/(2g)) Then substitute necessary values.

$${u}_{{x}} ={ucos}\theta;\:{u}_{{y}} ={usin}\theta;{u}=\mathrm{45}{ms}^{−\mathrm{1}} ;\:\theta=\mathrm{52}° \\ $$$${H}={maximum}\:{height};{R}={range} \\ $$$${t}={time}\:{from}\:{H}\:{to}\:\mathrm{12}{m}\:{above}\:{launching}\:{point}; \\ $$$${T}={flight}\:{time}\: \\ $$$${H}−\mathrm{12}=\frac{{gt}^{\mathrm{2}} }{\mathrm{2}}\Rightarrow\frac{{g}\left(\frac{{u}^{\mathrm{2}} {sin}^{\mathrm{2}} \theta}{{g}^{\mathrm{2}} }\right)}{\mathrm{2}}−\mathrm{12}=\frac{{gt}^{\mathrm{2}} }{\mathrm{2}} \\ $$$$\Rightarrow\frac{{u}^{\mathrm{2}} {sin}^{\mathrm{2}} \theta−\mathrm{24}{g}}{\mathrm{2}{g}}=\frac{{gt}^{\mathrm{2}} }{\mathrm{2}}\Rightarrow{t}=\frac{\sqrt{{u}^{\mathrm{2}} {sin}^{\mathrm{2}} \theta−\mathrm{24}{g}}}{{g}} \\ $$$$\Rightarrow{T}=\frac{\sqrt{{u}^{\mathrm{2}} {sin}^{\mathrm{2}} \theta−\mathrm{24}{g}}}{{g}}+\frac{{usin}\theta}{{g}} \\ $$$${R}={u}_{{x}} {T}=\frac{{ucos}\theta\sqrt{{u}^{\mathrm{2}} {sin}^{\mathrm{2}} \theta−\mathrm{24}{g}}}{{g}}+\frac{{u}^{\mathrm{2}} {sin}\mathrm{2}\theta}{\mathrm{2}{g}} \\ $$$${Then}\:{substitute}\:{necessary}\:{values}. \\ $$

Commented by Tawa11 last updated on 21/Nov/24

$$\mathrm{Thanks}\:\mathrm{sir}.\:\mathrm{I}\:\mathrm{appreciate}. \\ $$

Facebook Tweet Pin

Leave a Reply Cancel reply