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Question Number 18327    Answers: 1   Comments: 1

Question Number 18323    Answers: 0   Comments: 0

Σ((cos 2rθ)/(sin^2 2rθ−sin^2 θ))

$$\Sigma\frac{\mathrm{cos}\:\mathrm{2}{r}\theta}{\mathrm{sin}\:^{\mathrm{2}} \mathrm{2}{r}\theta−\mathrm{sin}\:^{\mathrm{2}} \theta} \\ $$

Question Number 18322    Answers: 1   Comments: 1

The pulley arrangements are identical. The mass of the rope is negligible. In (a), the mass m is lifted up by attaching a mass (2m) to the other end of the rope. In (b), m is lifted up by pulling the other end of the rope with a constant downward force F = 2mg. In which case, the acceleration of m is more?

$$\mathrm{The}\:\mathrm{pulley}\:\mathrm{arrangements}\:\mathrm{are}\:\mathrm{identical}. \\ $$$$\mathrm{The}\:\mathrm{mass}\:\mathrm{of}\:\mathrm{the}\:\mathrm{rope}\:\mathrm{is}\:\mathrm{negligible}.\:\mathrm{In} \\ $$$$\left(\mathrm{a}\right),\:\mathrm{the}\:\mathrm{mass}\:{m}\:\mathrm{is}\:\mathrm{lifted}\:\mathrm{up}\:\mathrm{by}\:\mathrm{attaching} \\ $$$$\mathrm{a}\:\mathrm{mass}\:\left(\mathrm{2}{m}\right)\:\mathrm{to}\:\mathrm{the}\:\mathrm{other}\:\mathrm{end}\:\mathrm{of}\:\mathrm{the}\:\mathrm{rope}. \\ $$$$\mathrm{In}\:\left(\mathrm{b}\right),\:{m}\:\mathrm{is}\:\mathrm{lifted}\:\mathrm{up}\:\mathrm{by}\:\mathrm{pulling}\:\mathrm{the} \\ $$$$\mathrm{other}\:\mathrm{end}\:\mathrm{of}\:\mathrm{the}\:\mathrm{rope}\:\mathrm{with}\:\mathrm{a}\:\mathrm{constant} \\ $$$$\mathrm{downward}\:\mathrm{force}\:{F}\:=\:\mathrm{2}{mg}.\:\mathrm{In}\:\mathrm{which} \\ $$$$\mathrm{case},\:\mathrm{the}\:\mathrm{acceleration}\:\mathrm{of}\:{m}\:\mathrm{is}\:\mathrm{more}? \\ $$

Question Number 18318    Answers: 0   Comments: 3

∫ ((x + sinx)/(cosx)) dx

$$\int\:\frac{\mathrm{x}\:+\:\mathrm{sinx}}{\mathrm{cosx}}\:\mathrm{dx} \\ $$

Question Number 18307    Answers: 1   Comments: 0

Question Number 18306    Answers: 1   Comments: 0

Question Number 18320    Answers: 1   Comments: 0

In a triangle ABC with fixed base BC, the vertex A moves such that cos B + cos C = 4 sin^2 (A/2) . If a, b and c denote the lengths of the sides of the triangle opposite to the angles A, B and C respectively, then (1) b + c = 4a (2) b + c = 2a (3) Locus of point A is an ellipse (4) Locus of point A is a pair of straight lines

$$\mathrm{In}\:\mathrm{a}\:\mathrm{triangle}\:{ABC}\:\mathrm{with}\:\mathrm{fixed}\:\mathrm{base}\:{BC}, \\ $$$$\mathrm{the}\:\mathrm{vertex}\:{A}\:\mathrm{moves}\:\mathrm{such}\:\mathrm{that}\:\mathrm{cos}\:{B}\:+ \\ $$$$\mathrm{cos}\:{C}\:=\:\mathrm{4}\:\mathrm{sin}^{\mathrm{2}} \:\frac{{A}}{\mathrm{2}}\:.\:\mathrm{If}\:{a},\:{b}\:\mathrm{and}\:{c}\:\mathrm{denote} \\ $$$$\mathrm{the}\:\mathrm{lengths}\:\mathrm{of}\:\mathrm{the}\:\mathrm{sides}\:\mathrm{of}\:\mathrm{the}\:\mathrm{triangle} \\ $$$$\mathrm{opposite}\:\mathrm{to}\:\mathrm{the}\:\mathrm{angles}\:{A},\:{B}\:\mathrm{and}\:{C} \\ $$$$\mathrm{respectively},\:\mathrm{then} \\ $$$$\left(\mathrm{1}\right)\:{b}\:+\:{c}\:=\:\mathrm{4}{a} \\ $$$$\left(\mathrm{2}\right)\:{b}\:+\:{c}\:=\:\mathrm{2}{a} \\ $$$$\left(\mathrm{3}\right)\:\mathrm{Locus}\:\mathrm{of}\:\mathrm{point}\:{A}\:\mathrm{is}\:\mathrm{an}\:\mathrm{ellipse} \\ $$$$\left(\mathrm{4}\right)\:\mathrm{Locus}\:\mathrm{of}\:\mathrm{point}\:{A}\:\mathrm{is}\:\mathrm{a}\:\mathrm{pair}\:\mathrm{of}\:\mathrm{straight} \\ $$$$\mathrm{lines} \\ $$

Question Number 18719    Answers: 0   Comments: 2

If cos^2 x_1 + cos^2 x_2 + cos^2 x_3 + cos^2 x_4 + cos^2 x_5 = 5, then sin x_1 + 2sin x_2 + 3sin x_3 + 4sin x_4 + 5sin x_5 is less than or equal to

$$\mathrm{If}\:\mathrm{cos}^{\mathrm{2}} \:{x}_{\mathrm{1}} \:+\:\mathrm{cos}^{\mathrm{2}} \:{x}_{\mathrm{2}} \:+\:\mathrm{cos}^{\mathrm{2}} \:{x}_{\mathrm{3}} \:+\:\mathrm{cos}^{\mathrm{2}} \:{x}_{\mathrm{4}} \\ $$$$+\:\mathrm{cos}^{\mathrm{2}} \:{x}_{\mathrm{5}} \:=\:\mathrm{5},\:\mathrm{then}\:\mathrm{sin}\:{x}_{\mathrm{1}} \:+\:\mathrm{2sin}\:{x}_{\mathrm{2}} \:+ \\ $$$$\mathrm{3sin}\:{x}_{\mathrm{3}} \:+\:\mathrm{4sin}\:{x}_{\mathrm{4}} \:+\:\mathrm{5sin}\:{x}_{\mathrm{5}} \:\mathrm{is}\:\mathrm{less}\:\mathrm{than} \\ $$$$\mathrm{or}\:\mathrm{equal}\:\mathrm{to} \\ $$

Question Number 18299    Answers: 0   Comments: 2

x^x^x = 2, find x

$$\mathrm{x}^{\mathrm{x}^{\mathrm{x}} } \:=\:\mathrm{2},\:\:\:\:\:\:\mathrm{find}\:\:\mathrm{x} \\ $$

Question Number 18470    Answers: 1   Comments: 0

Find the partial derivatives for each of the following (a) Z = 3x^2 (5x + 7y)^2 (b) Z = (w − x − y)^2 (3w + 2x − 4y)

$$\mathrm{Find}\:\mathrm{the}\:\mathrm{partial}\:\mathrm{derivatives}\:\mathrm{for}\:\mathrm{each}\:\mathrm{of}\:\mathrm{the}\:\mathrm{following} \\ $$$$\left(\mathrm{a}\right)\:\mathrm{Z}\:=\:\mathrm{3x}^{\mathrm{2}} \left(\mathrm{5x}\:+\:\mathrm{7y}\right)^{\mathrm{2}} \\ $$$$\left(\mathrm{b}\right)\:\mathrm{Z}\:=\:\left(\mathrm{w}\:−\:\mathrm{x}\:−\:\mathrm{y}\right)^{\mathrm{2}} \:\left(\mathrm{3w}\:+\:\mathrm{2x}\:−\:\mathrm{4y}\right) \\ $$

Question Number 18290    Answers: 0   Comments: 3

Question Number 18285    Answers: 0   Comments: 1

∫_0 ^5 (1/(∫_1 ^8 e^x^(−5) ))dx

$$\int_{\mathrm{0}} ^{\mathrm{5}} \frac{\mathrm{1}}{\int_{\mathrm{1}} ^{\mathrm{8}} \mathrm{e}^{\mathrm{x}^{−\mathrm{5}} } }\mathrm{dx} \\ $$

Question Number 18284    Answers: 1   Comments: 0

If k is any possible number. what is the size of angle between the vectors a(K, k) and b(− 3, 4)

$$\mathrm{If}\:\mathrm{k}\:\mathrm{is}\:\mathrm{any}\:\mathrm{possible}\:\mathrm{number}.\:\mathrm{what}\:\mathrm{is}\:\mathrm{the}\:\mathrm{size}\:\mathrm{of}\:\mathrm{angle}\:\mathrm{between}\:\mathrm{the}\:\mathrm{vectors} \\ $$$$\mathrm{a}\left(\mathrm{K},\:\mathrm{k}\right)\:\mathrm{and}\:\mathrm{b}\left(−\:\mathrm{3},\:\mathrm{4}\right) \\ $$

Question Number 18283    Answers: 0   Comments: 0

Given: x^2 + y^2 Show that, (d^2 y/dx^2 ) = ((xy)/(y^2 + x^2 ))

$$\mathrm{Given}:\:\:\mathrm{x}^{\mathrm{2}} \:+\:\mathrm{y}^{\mathrm{2}} \\ $$$$\mathrm{Show}\:\mathrm{that},\:\:\:\frac{\mathrm{d}^{\mathrm{2}} \mathrm{y}}{\mathrm{dx}^{\mathrm{2}} }\:=\:\frac{\mathrm{xy}}{\mathrm{y}^{\mathrm{2}} \:+\:\mathrm{x}^{\mathrm{2}} } \\ $$

Question Number 18282    Answers: 0   Comments: 0

Two trains 150 m long and 250 m long are travelling at the speed of 30 kmph and 33 kmph respectively on parallel tracks in opposite directions. What is the time taken by these trains to cross each other completely from the moment they meet?

$$\mathrm{Two}\:\mathrm{trains}\:\mathrm{150}\:\mathrm{m}\:\mathrm{long}\:\mathrm{and}\:\mathrm{250}\:\mathrm{m}\:\mathrm{long} \\ $$$$\mathrm{are}\:\mathrm{travelling}\:\mathrm{at}\:\mathrm{the}\:\mathrm{speed}\:\mathrm{of}\:\mathrm{30}\:\mathrm{kmph} \\ $$$$\mathrm{and}\:\mathrm{33}\:\mathrm{kmph}\:\mathrm{respectively}\:\mathrm{on}\:\mathrm{parallel} \\ $$$$\mathrm{tracks}\:\mathrm{in}\:\mathrm{opposite}\:\mathrm{directions}.\:\mathrm{What}\:\mathrm{is} \\ $$$$\mathrm{the}\:\mathrm{time}\:\mathrm{taken}\:\mathrm{by}\:\mathrm{these}\:\mathrm{trains}\:\mathrm{to}\:\mathrm{cross} \\ $$$$\mathrm{each}\:\mathrm{other}\:\mathrm{completely}\:\mathrm{from}\:\mathrm{the} \\ $$$$\mathrm{moment}\:\mathrm{they}\:\mathrm{meet}? \\ $$

Question Number 18281    Answers: 2   Comments: 0

Question Number 18278    Answers: 2   Comments: 1

Question Number 18279    Answers: 0   Comments: 0

In an 1800 m race, P beats Q by 50 seconds. In the same race, Q beats R by 40 seconds. If P beats R by 450 m, by what distance does P beat Q ?(in m)

$$\mathrm{In}\:\mathrm{an}\:\mathrm{1800}\:\mathrm{m}\:\mathrm{race},\:\mathrm{P}\:\mathrm{beats}\:\mathrm{Q}\:\mathrm{by}\:\mathrm{50}\: \\ $$$$\mathrm{seconds}.\:\mathrm{In}\:\mathrm{the}\:\mathrm{same}\:\mathrm{race},\:\mathrm{Q}\:\mathrm{beats}\:\mathrm{R}\:\mathrm{by} \\ $$$$\mathrm{40}\:\mathrm{seconds}.\:\mathrm{If}\:\mathrm{P}\:\mathrm{beats}\:\mathrm{R}\:\mathrm{by}\:\mathrm{450}\:\mathrm{m},\:\mathrm{by}\: \\ $$$$\mathrm{what}\:\mathrm{distance}\:\mathrm{does}\:\mathrm{P}\:\mathrm{beat}\:\mathrm{Q}\:?\left(\mathrm{in}\:\mathrm{m}\right) \\ $$

Question Number 18274    Answers: 1   Comments: 0

A balloon moves up vertically such that if a stone is projected with a horizontal velocity u relative to balloon, the stone always hits the ground at a fixed point at a distance ((2u^2 )/g) horizontally away from it. Find the height of the balloon as a function of time.

$$\mathrm{A}\:\mathrm{balloon}\:\mathrm{moves}\:\mathrm{up}\:\mathrm{vertically}\:\mathrm{such} \\ $$$$\mathrm{that}\:\mathrm{if}\:\mathrm{a}\:\mathrm{stone}\:\mathrm{is}\:\mathrm{projected}\:\mathrm{with}\:\mathrm{a} \\ $$$$\mathrm{horizontal}\:\mathrm{velocity}\:{u}\:\mathrm{relative}\:\mathrm{to}\:\mathrm{balloon}, \\ $$$$\mathrm{the}\:\mathrm{stone}\:\mathrm{always}\:\mathrm{hits}\:\mathrm{the}\:\mathrm{ground}\:\mathrm{at}\:\mathrm{a} \\ $$$$\mathrm{fixed}\:\mathrm{point}\:\mathrm{at}\:\mathrm{a}\:\mathrm{distance}\:\frac{\mathrm{2}{u}^{\mathrm{2}} }{{g}}\:\mathrm{horizontally} \\ $$$$\mathrm{away}\:\mathrm{from}\:\mathrm{it}.\:\mathrm{Find}\:\mathrm{the}\:\mathrm{height}\:\mathrm{of}\:\mathrm{the} \\ $$$$\mathrm{balloon}\:\mathrm{as}\:\mathrm{a}\:\mathrm{function}\:\mathrm{of}\:\mathrm{time}. \\ $$

Question Number 18271    Answers: 0   Comments: 3

There are two parallel planes, each inclined to the horizontal at an angle θ. A particle is projected from a point mid way between the foot of the two planes so that it grazes one of the planes and strikes the other at right angle. Find the angle of projection of the projectile.

$$\mathrm{There}\:\mathrm{are}\:\mathrm{two}\:\mathrm{parallel}\:\mathrm{planes},\:\mathrm{each} \\ $$$$\mathrm{inclined}\:\mathrm{to}\:\mathrm{the}\:\mathrm{horizontal}\:\mathrm{at}\:\mathrm{an}\:\mathrm{angle}\:\theta. \\ $$$$\mathrm{A}\:\mathrm{particle}\:\mathrm{is}\:\mathrm{projected}\:\mathrm{from}\:\mathrm{a}\:\mathrm{point}\:\mathrm{mid} \\ $$$$\mathrm{way}\:\mathrm{between}\:\mathrm{the}\:\mathrm{foot}\:\mathrm{of}\:\mathrm{the}\:\mathrm{two}\:\mathrm{planes} \\ $$$$\mathrm{so}\:\mathrm{that}\:\mathrm{it}\:\mathrm{grazes}\:\mathrm{one}\:\mathrm{of}\:\mathrm{the}\:\mathrm{planes}\:\mathrm{and} \\ $$$$\mathrm{strikes}\:\mathrm{the}\:\mathrm{other}\:\mathrm{at}\:\mathrm{right}\:\mathrm{angle}.\:\mathrm{Find} \\ $$$$\mathrm{the}\:\mathrm{angle}\:\mathrm{of}\:\mathrm{projection}\:\mathrm{of}\:\mathrm{the}\:\mathrm{projectile}. \\ $$

Question Number 18269    Answers: 0   Comments: 3

The flow velocity of a river increases linearly with the distance (r) from its bank and has its maximum value v_0 in the middle of the river. The velocity near the bank is zero. A boat which can move with speed u in still water moves in the river in such a way that it is always perpendicular to the flow of current. Find (i) The distance along the bank through which boat is carried away by the flow current, when the boat crosses the river. (ii) The equation of trajectory for the coordinate system shown. Assume that the swimmer starts from origin.

$$\mathrm{The}\:\mathrm{flow}\:\mathrm{velocity}\:\mathrm{of}\:\mathrm{a}\:\mathrm{river}\:\mathrm{increases} \\ $$$$\mathrm{linearly}\:\mathrm{with}\:\mathrm{the}\:\mathrm{distance}\:\left({r}\right)\:\mathrm{from}\:\mathrm{its} \\ $$$$\mathrm{bank}\:\mathrm{and}\:\mathrm{has}\:\mathrm{its}\:\mathrm{maximum}\:\mathrm{value}\:{v}_{\mathrm{0}} \:\mathrm{in} \\ $$$$\mathrm{the}\:\mathrm{middle}\:\mathrm{of}\:\mathrm{the}\:\mathrm{river}.\:\mathrm{The}\:\mathrm{velocity} \\ $$$$\mathrm{near}\:\mathrm{the}\:\mathrm{bank}\:\mathrm{is}\:\mathrm{zero}.\:\mathrm{A}\:\mathrm{boat}\:\mathrm{which}\:\mathrm{can} \\ $$$$\mathrm{move}\:\mathrm{with}\:\mathrm{speed}\:{u}\:\mathrm{in}\:\mathrm{still}\:\mathrm{water}\:\mathrm{moves} \\ $$$$\mathrm{in}\:\mathrm{the}\:\mathrm{river}\:\mathrm{in}\:\mathrm{such}\:\mathrm{a}\:\mathrm{way}\:\mathrm{that}\:\mathrm{it}\:\mathrm{is} \\ $$$$\mathrm{always}\:\mathrm{perpendicular}\:\mathrm{to}\:\mathrm{the}\:\mathrm{flow}\:\mathrm{of} \\ $$$$\mathrm{current}.\:\mathrm{Find} \\ $$$$\left(\mathrm{i}\right)\:\mathrm{The}\:\mathrm{distance}\:\mathrm{along}\:\mathrm{the}\:\mathrm{bank}\:\mathrm{through} \\ $$$$\mathrm{which}\:\mathrm{boat}\:\mathrm{is}\:\mathrm{carried}\:\mathrm{away}\:\mathrm{by}\:\mathrm{the}\:\mathrm{flow} \\ $$$$\mathrm{current},\:\mathrm{when}\:\mathrm{the}\:\mathrm{boat}\:\mathrm{crosses}\:\mathrm{the} \\ $$$$\mathrm{river}. \\ $$$$\left(\mathrm{ii}\right)\:\mathrm{The}\:\mathrm{equation}\:\mathrm{of}\:\mathrm{trajectory}\:\mathrm{for}\:\mathrm{the} \\ $$$$\mathrm{coordinate}\:\mathrm{system}\:\mathrm{shown}.\:\mathrm{Assume} \\ $$$$\mathrm{that}\:\mathrm{the}\:\mathrm{swimmer}\:\mathrm{starts}\:\mathrm{from}\:\mathrm{origin}. \\ $$

Question Number 18268    Answers: 1   Comments: 0

A balloon starts rising from the surface of earth. The ascension rate is constant and is equal to v_0 . Due to wind the balloon gathers horizontal velocity component v_x = ay, where a is a positive constant and y is the height of ascent. Find (i) The horizontal drift of the balloon x(y), (ii) The total, tangential and normal accelerations of the balloon.

$$\mathrm{A}\:\mathrm{balloon}\:\mathrm{starts}\:\mathrm{rising}\:\mathrm{from}\:\mathrm{the}\:\mathrm{surface} \\ $$$$\mathrm{of}\:\mathrm{earth}.\:\mathrm{The}\:\mathrm{ascension}\:\mathrm{rate}\:\mathrm{is}\:\mathrm{constant} \\ $$$$\mathrm{and}\:\mathrm{is}\:\mathrm{equal}\:\mathrm{to}\:{v}_{\mathrm{0}} .\:\mathrm{Due}\:\mathrm{to}\:\mathrm{wind}\:\mathrm{the} \\ $$$$\mathrm{balloon}\:\mathrm{gathers}\:\mathrm{horizontal}\:\mathrm{velocity} \\ $$$$\mathrm{component}\:{v}_{{x}} \:=\:{ay},\:\mathrm{where}\:{a}\:\mathrm{is}\:\mathrm{a}\:\mathrm{positive} \\ $$$$\mathrm{constant}\:\mathrm{and}\:{y}\:\mathrm{is}\:\mathrm{the}\:\mathrm{height}\:\mathrm{of}\:\mathrm{ascent}. \\ $$$$\mathrm{Find} \\ $$$$\left(\mathrm{i}\right)\:\mathrm{The}\:\mathrm{horizontal}\:\mathrm{drift}\:\mathrm{of}\:\mathrm{the}\:\mathrm{balloon} \\ $$$${x}\left({y}\right), \\ $$$$\left(\mathrm{ii}\right)\:\mathrm{The}\:\mathrm{total},\:\mathrm{tangential}\:\mathrm{and}\:\mathrm{normal} \\ $$$$\mathrm{accelerations}\:\mathrm{of}\:\mathrm{the}\:\mathrm{balloon}. \\ $$

Question Number 19654    Answers: 0   Comments: 3

Two swimmers leave point A on one bank of the river to reach point B lying right across the other bank. One of them crosses the river along the straight line AB while the other swims at right angle to the stream and then walks the distance that he has been carried away by the stream to get to point B. What was the velocity v of his walking if both swimmers reached the destination simultaneously? (The stream velocity v_0 = 2 km/h and the velocity v′ of each swimmer with respect to still water is 2.5 km/h).

$$\mathrm{Two}\:\mathrm{swimmers}\:\mathrm{leave}\:\mathrm{point}\:{A}\:\mathrm{on}\:\mathrm{one} \\ $$$$\mathrm{bank}\:\mathrm{of}\:\mathrm{the}\:\mathrm{river}\:\mathrm{to}\:\mathrm{reach}\:\mathrm{point}\:{B}\:\mathrm{lying} \\ $$$$\mathrm{right}\:\mathrm{across}\:\mathrm{the}\:\mathrm{other}\:\mathrm{bank}.\:\mathrm{One}\:\mathrm{of} \\ $$$$\mathrm{them}\:\mathrm{crosses}\:\mathrm{the}\:\mathrm{river}\:\mathrm{along}\:\mathrm{the}\:\mathrm{straight} \\ $$$$\mathrm{line}\:{AB}\:\mathrm{while}\:\mathrm{the}\:\mathrm{other}\:\mathrm{swims}\:\mathrm{at}\:\mathrm{right} \\ $$$$\mathrm{angle}\:\mathrm{to}\:\mathrm{the}\:\mathrm{stream}\:\mathrm{and}\:\mathrm{then}\:\mathrm{walks}\:\mathrm{the} \\ $$$$\mathrm{distance}\:\mathrm{that}\:\mathrm{he}\:\mathrm{has}\:\mathrm{been}\:\mathrm{carried}\:\mathrm{away} \\ $$$$\mathrm{by}\:\mathrm{the}\:\mathrm{stream}\:\mathrm{to}\:\mathrm{get}\:\mathrm{to}\:\mathrm{point}\:{B}.\:\mathrm{What} \\ $$$$\mathrm{was}\:\mathrm{the}\:\mathrm{velocity}\:{v}\:\mathrm{of}\:\mathrm{his}\:\mathrm{walking}\:\mathrm{if}\:\mathrm{both} \\ $$$$\mathrm{swimmers}\:\mathrm{reached}\:\mathrm{the}\:\mathrm{destination} \\ $$$$\mathrm{simultaneously}?\:\left(\mathrm{The}\:\mathrm{stream}\:\mathrm{velocity}\right. \\ $$$${v}_{\mathrm{0}} \:=\:\mathrm{2}\:\mathrm{km}/\mathrm{h}\:\mathrm{and}\:\mathrm{the}\:\mathrm{velocity}\:{v}'\:\mathrm{of}\:\mathrm{each} \\ $$$$\mathrm{swimmer}\:\mathrm{with}\:\mathrm{respect}\:\mathrm{to}\:\mathrm{still}\:\mathrm{water}\:\mathrm{is} \\ $$$$\left.\mathrm{2}.\mathrm{5}\:\mathrm{km}/\mathrm{h}\right). \\ $$

Question Number 18265    Answers: 0   Comments: 7

A particle is projected at an angle 60° with speed 10(√3) m/s from the point A as shown in the figure. At the same time the wedge is made to move with speed 10(√3) m/s toward right as shown in figure. Find the time after which particle will strike the wedge.

$$\mathrm{A}\:\mathrm{particle}\:\mathrm{is}\:\mathrm{projected}\:\mathrm{at}\:\mathrm{an}\:\mathrm{angle}\:\mathrm{60}° \\ $$$$\mathrm{with}\:\mathrm{speed}\:\mathrm{10}\sqrt{\mathrm{3}}\:\mathrm{m}/\mathrm{s}\:\mathrm{from}\:\mathrm{the}\:\mathrm{point}\:{A} \\ $$$$\mathrm{as}\:\mathrm{shown}\:\mathrm{in}\:\mathrm{the}\:\mathrm{figure}.\:\mathrm{At}\:\mathrm{the}\:\mathrm{same} \\ $$$$\mathrm{time}\:\mathrm{the}\:\mathrm{wedge}\:\mathrm{is}\:\mathrm{made}\:\mathrm{to}\:\mathrm{move}\:\mathrm{with} \\ $$$$\mathrm{speed}\:\mathrm{10}\sqrt{\mathrm{3}}\:\mathrm{m}/\mathrm{s}\:\mathrm{toward}\:\mathrm{right}\:\mathrm{as}\:\mathrm{shown} \\ $$$$\mathrm{in}\:\mathrm{figure}.\:\mathrm{Find}\:\mathrm{the}\:\mathrm{time}\:\mathrm{after}\:\mathrm{which} \\ $$$$\mathrm{particle}\:\mathrm{will}\:\mathrm{strike}\:\mathrm{the}\:\mathrm{wedge}. \\ $$

Question Number 18264    Answers: 1   Comments: 0

A sky diver of mass m drops out with an initial velocity v_0 = 0. Find the law by which the sky diver′s speed varies before the parachute is opened if the drag is proportional to the sky diver′s speed. Also solve the problem when the sky diver′s initial velocity has horizontal component v_0 and vertical component zero.

$$\mathrm{A}\:\mathrm{sky}\:\mathrm{diver}\:\mathrm{of}\:\mathrm{mass}\:{m}\:\mathrm{drops}\:\mathrm{out}\:\mathrm{with} \\ $$$$\mathrm{an}\:\mathrm{initial}\:\mathrm{velocity}\:{v}_{\mathrm{0}} \:=\:\mathrm{0}.\:\mathrm{Find}\:\mathrm{the}\:\mathrm{law} \\ $$$$\mathrm{by}\:\mathrm{which}\:\mathrm{the}\:\mathrm{sky}\:\mathrm{diver}'\mathrm{s}\:\mathrm{speed}\:\mathrm{varies} \\ $$$$\mathrm{before}\:\mathrm{the}\:\mathrm{parachute}\:\mathrm{is}\:\mathrm{opened}\:\mathrm{if}\:\mathrm{the} \\ $$$$\mathrm{drag}\:\mathrm{is}\:\mathrm{proportional}\:\mathrm{to}\:\mathrm{the}\:\mathrm{sky}\:\mathrm{diver}'\mathrm{s} \\ $$$$\mathrm{speed}.\:\mathrm{Also}\:\mathrm{solve}\:\mathrm{the}\:\mathrm{problem}\:\mathrm{when}\:\mathrm{the} \\ $$$$\mathrm{sky}\:\mathrm{diver}'\mathrm{s}\:\mathrm{initial}\:\mathrm{velocity}\:\mathrm{has}\:\mathrm{horizontal} \\ $$$$\mathrm{component}\:{v}_{\mathrm{0}} \:\mathrm{and}\:\mathrm{vertical}\:\mathrm{component} \\ $$$$\mathrm{zero}. \\ $$

Question Number 18262    Answers: 1   Comments: 2

A point P is located above an inclined plane. It is possible to reach the plane by sliding under gravity down a straight frictionless wire joining to some point P ′ on the plane. How should P ′ be chosen so as to minimize the time taken?

$$\mathrm{A}\:\mathrm{point}\:{P}\:\mathrm{is}\:\mathrm{located}\:\mathrm{above}\:\mathrm{an}\:\mathrm{inclined} \\ $$$$\mathrm{plane}.\:\mathrm{It}\:\mathrm{is}\:\mathrm{possible}\:\mathrm{to}\:\mathrm{reach}\:\mathrm{the}\:\mathrm{plane} \\ $$$$\mathrm{by}\:\mathrm{sliding}\:\mathrm{under}\:\mathrm{gravity}\:\mathrm{down}\:\mathrm{a}\:\mathrm{straight} \\ $$$$\mathrm{frictionless}\:\mathrm{wire}\:\mathrm{joining}\:\mathrm{to}\:\mathrm{some}\:\mathrm{point} \\ $$$${P}\:'\:\mathrm{on}\:\mathrm{the}\:\mathrm{plane}.\:\mathrm{How}\:\mathrm{should}\:{P}\:'\:\mathrm{be} \\ $$$$\mathrm{chosen}\:\mathrm{so}\:\mathrm{as}\:\mathrm{to}\:\mathrm{minimize}\:\mathrm{the}\:\mathrm{time} \\ $$$$\mathrm{taken}? \\ $$

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