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Question Number 15932 by Tinkutara last updated on 15/Jun/17

A particle is projected from the foot of an inclined plane having inclination 45°, with the velocity u at an angle θ (> 45°) with the horizontal in a vertical plane containing the line of greatest slope through the point of projection. Find the value of tan θ if the particle strikes the plane (i) Horizontally (ii) Normally

$$\mathrm{A}\:\mathrm{particle}\:\mathrm{is}\:\mathrm{projected}\:\mathrm{from}\:\mathrm{the}\:\mathrm{foot}\:\mathrm{of} \\ $$ $$\mathrm{an}\:\mathrm{inclined}\:\mathrm{plane}\:\mathrm{having}\:\mathrm{inclination} \\ $$ $$\mathrm{45}°,\:\mathrm{with}\:\mathrm{the}\:\mathrm{velocity}\:{u}\:\mathrm{at}\:\mathrm{an}\:\mathrm{angle}\:\theta \\ $$ $$\left(>\:\mathrm{45}°\right)\:\mathrm{with}\:\mathrm{the}\:\mathrm{horizontal}\:\mathrm{in}\:\mathrm{a}\:\mathrm{vertical} \\ $$ $$\mathrm{plane}\:\mathrm{containing}\:\mathrm{the}\:\mathrm{line}\:\mathrm{of}\:\mathrm{greatest} \\ $$ $$\mathrm{slope}\:\mathrm{through}\:\mathrm{the}\:\mathrm{point}\:\mathrm{of}\:\mathrm{projection}. \\ $$ $$\mathrm{Find}\:\mathrm{the}\:\mathrm{value}\:\mathrm{of}\:\mathrm{tan}\:\theta\:\mathrm{if}\:\mathrm{the}\:\mathrm{particle} \\ $$ $$\mathrm{strikes}\:\mathrm{the}\:\mathrm{plane} \\ $$ $$\left(\mathrm{i}\right)\:\mathrm{Horizontally} \\ $$ $$\left(\mathrm{ii}\right)\:\mathrm{Normally} \\ $$

Commented byTinkutara last updated on 15/Jun/17

Answered by ajfour last updated on 15/Jun/17

when it hits the incline horizontally it is at its max. hright. tan α=(H/((R/2))) , and witb α=45° R = 2H ((2u^2 sin θcos θ)/g)=((2u^2 sin^2 θ)/(2g)) ⇒ tan θ = 2 or θ=tan^(−1) (2) . when it strikes nirmally, its component of velocity parallel to incline becomes zero by the time its displacement perpendicular to incline becomes zero. v_(x′) =ucos (θ−α)−gtsin α=0 or t=((ucos (θ−α))/(gsin α)) .....(i) s_(y′) =utsin (θ−α)−(1/2)gt^2 cos α=0 ⇒ t=((2usin (θ−α))/(gcos α)) ...(ii) equating (ii) and (i): ((2usin (θ−α))/(gcos α))=((ucos (θ−α))/(gsin α)) ⇒ tan (θ−α)tan α =(1/2) as α=45°, tan (θ−45°)×1=(1/2) θ= 45°+tan^(−1) ((1/2)) .

$${when}\:{it}\:{hits}\:{the}\:{incline}\:{horizontally} \\ $$ $${it}\:{is}\:{at}\:{its}\:{max}.\:{hright}. \\ $$ $$\mathrm{tan}\:\alpha=\frac{{H}}{\left({R}/\mathrm{2}\right)}\:\:\:\:,\:{and}\:{witb}\:\alpha=\mathrm{45}° \\ $$ $$\:{R}\:=\:\mathrm{2}{H} \\ $$ $$\:\frac{\mathrm{2}{u}^{\mathrm{2}} \mathrm{sin}\:\theta\mathrm{cos}\:\theta}{{g}}=\frac{\mathrm{2}{u}^{\mathrm{2}} \mathrm{sin}\:^{\mathrm{2}} \theta}{\mathrm{2}{g}} \\ $$ $$\Rightarrow\:\:\mathrm{tan}\:\theta\:=\:\mathrm{2}\:\:{or}\:\:\theta=\mathrm{tan}^{−\mathrm{1}} \left(\mathrm{2}\right)\:. \\ $$ $${when}\:{it}\:{strikes}\:{nirmally}, \\ $$ $${its}\:{component}\:{of}\:{velocity}\: \\ $$ $${parallel}\:{to}\:{incline}\:{becomes}\:{zero} \\ $$ $${by}\:{the}\:{time}\:{its}\:{displacement} \\ $$ $${perpendicular}\:{to}\:{incline}\:{becomes} \\ $$ $${zero}. \\ $$ $$\:{v}_{{x}'} ={u}\mathrm{cos}\:\left(\theta−\alpha\right)−{gt}\mathrm{sin}\:\alpha=\mathrm{0} \\ $$ $${or}\:\:{t}=\frac{{u}\mathrm{cos}\:\left(\theta−\alpha\right)}{{g}\mathrm{sin}\:\alpha}\:\:\:\:\:\:.....\left({i}\right) \\ $$ $$\:{s}_{{y}'} ={ut}\mathrm{sin}\:\left(\theta−\alpha\right)−\frac{\mathrm{1}}{\mathrm{2}}{gt}^{\mathrm{2}} \mathrm{cos}\:\alpha=\mathrm{0} \\ $$ $$\Rightarrow\:{t}=\frac{\mathrm{2}{u}\mathrm{sin}\:\left(\theta−\alpha\right)}{{g}\mathrm{cos}\:\alpha}\:\:\:\:\:\:\:\:...\left({ii}\right) \\ $$ $${equating}\:\left({ii}\right)\:{and}\:\left({i}\right): \\ $$ $$\:\frac{\mathrm{2}{u}\mathrm{sin}\:\left(\theta−\alpha\right)}{{g}\mathrm{cos}\:\alpha}=\frac{{u}\mathrm{cos}\:\left(\theta−\alpha\right)}{{g}\mathrm{sin}\:\alpha} \\ $$ $$\Rightarrow\:\:\:\mathrm{tan}\:\left(\theta−\alpha\right)\mathrm{tan}\:\alpha\:=\frac{\mathrm{1}}{\mathrm{2}} \\ $$ $$\:\:{as}\:\alpha=\mathrm{45}°,\:\:\:\mathrm{tan}\:\left(\theta−\mathrm{45}°\right)×\mathrm{1}=\frac{\mathrm{1}}{\mathrm{2}} \\ $$ $$\:\theta=\:\mathrm{45}°+\mathrm{tan}^{−\mathrm{1}} \left(\frac{\mathrm{1}}{\mathrm{2}}\right)\:\:. \\ $$

Commented byTinkutara last updated on 16/Jun/17

Thanks Sir!

$$\mathrm{Thanks}\:\mathrm{Sir}! \\ $$

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