Question and Answers Forum

A particle P of mass m, is projected vertically upward with a speed u from a point A, on horizontal ground. When P is at x above its initial position, its speed is v. The only forces acting on P is its weight and resistance mgkv^2 . where k is a positive constant. (a) Show that the greatest height reached is (1/(2gk)) ln(1 +ku^2 ). (b) show that the speed with which P returns to A is (u/(√(1+ ku^2 ))) .

$$\mathrm{A}\:\mathrm{particle}\:\mathrm{P}\:\mathrm{of}\:\mathrm{mass}\:{m},\:\mathrm{is}\:\mathrm{projected}\:\mathrm{vertically}\:\mathrm{upward}\:\mathrm{with} \\ $$$$\mathrm{a}\:\mathrm{speed}\:{u}\:\mathrm{from}\:\mathrm{a}\:\mathrm{point}\:{A},\:\mathrm{on}\:\mathrm{horizontal}\:\mathrm{ground}.\:\mathrm{When}\:\mathrm{P}\:\mathrm{is}\:\mathrm{at}\:{x}\:\mathrm{above} \\ $$$$\mathrm{its}\:\mathrm{initial}\:\mathrm{position},\:\mathrm{its}\:\mathrm{speed}\:\mathrm{is}\:{v}.\:\mathrm{The}\:\mathrm{only}\:\mathrm{forces}\:\mathrm{acting}\:\mathrm{on}\:\mathrm{P}\:\mathrm{is} \\ $$$$\mathrm{its}\:\mathrm{weight}\:\mathrm{and}\:\mathrm{resistance}\:{m}\mathrm{g}{kv}^{\mathrm{2}} .\:\mathrm{where}\:{k}\:\mathrm{is}\:\mathrm{a}\:\mathrm{positive}\:\mathrm{constant}. \\ $$$$\left(\mathrm{a}\right)\:\mathrm{Show}\:\mathrm{that}\:\mathrm{the}\:\mathrm{greatest}\:\mathrm{height}\:\mathrm{reached}\:\mathrm{is}\:\frac{\mathrm{1}}{\mathrm{2g}{k}}\:\mathrm{ln}\left(\mathrm{1}\:+{ku}^{\mathrm{2}} \right). \\ $$$$\left(\mathrm{b}\right)\:\mathrm{show}\:\mathrm{that}\:\mathrm{the}\:\mathrm{speed}\:\mathrm{with}\:\mathrm{which}\:\mathrm{P}\:\mathrm{returns}\:\mathrm{to}\:\mathrm{A}\:\mathrm{is}\:\frac{{u}}{\sqrt{\mathrm{1}+\:{ku}^{\mathrm{2}} }}\:. \\ $$

Question Number 96763 Answers: 2 Comments: 0

∫ ((sin ((x/2)) tan ((x/2)) dx)/(cos x)) = ?

$$\int\:\frac{\mathrm{sin}\:\left(\frac{\mathrm{x}}{\mathrm{2}}\right)\:\mathrm{tan}\:\left(\frac{\mathrm{x}}{\mathrm{2}}\right)\:\mathrm{dx}}{\mathrm{cos}\:\mathrm{x}}\:=\:? \\ $$

Question Number 96761 Answers: 0 Comments: 0

A particle P moving at constant angular velocity describes a part y = f(θ). At time t = 0, the particle is at the point with coordinate (a,(π/2)) and moving with a transverse acceleration of −2aω^2 sinθ. find the polar equation of the curve described by this particle.Show that the radial component of the acceleration of P is −aω^2 (1 + cos θ).

$$\mathrm{A}\:\mathrm{particle}\:\mathrm{P}\:\:\:\mathrm{moving}\:\mathrm{at}\:\mathrm{constant}\:\mathrm{angular}\:\mathrm{velocity} \\ $$$$\mathrm{describes}\:\mathrm{a}\:\mathrm{part}\:{y}\:=\:{f}\left(\theta\right).\:\mathrm{At}\:\mathrm{time}\:{t}\:=\:\mathrm{0},\:\mathrm{the}\:\mathrm{particle} \\ $$$$\mathrm{is}\:\mathrm{at}\:\mathrm{the}\:\mathrm{point}\:\mathrm{with}\:\mathrm{coordinate}\:\left({a},\frac{\pi}{\mathrm{2}}\right)\:\mathrm{and}\:\mathrm{moving}\:\mathrm{with}\:\mathrm{a}\: \\ $$$$\mathrm{transverse}\:\mathrm{acceleration}\:\mathrm{of}\:−\mathrm{2}{a}\omega^{\mathrm{2}} \:\mathrm{sin}\theta.\:\mathrm{find}\:\mathrm{the}\:\mathrm{polar}\:\mathrm{equation} \\ $$$$\mathrm{of}\:\mathrm{the}\:\mathrm{curve}\:\mathrm{described}\:\mathrm{by}\:\mathrm{this}\:\mathrm{particle}.\mathrm{Show}\:\mathrm{that}\:\mathrm{the} \\ $$$$\mathrm{radial}\:\mathrm{component}\:\mathrm{of}\:\mathrm{the}\:\:\mathrm{acceleration}\:\:\mathrm{of}\:\mathrm{P}\:\mathrm{is}\:−{a}\omega^{\mathrm{2}} \left(\mathrm{1}\:+\:\mathrm{cos}\:\theta\right). \\ $$

Question Number 96758 Answers: 0 Comments: 1

Let x∈ [ −((5π)/(12)) , −(π/3) ] . The maximum value of y = tan (x+((2π)/3))−tan (x+(π/6)) +cos (x+(π/6)) is ___

$$\mathrm{Let}\:{x}\in\:\left[\:−\frac{\mathrm{5}\pi}{\mathrm{12}}\:,\:−\frac{\pi}{\mathrm{3}}\:\right]\:.\:\mathrm{The}\:\mathrm{maximum}\: \\ $$$$\mathrm{value}\:\mathrm{of}\:\mathrm{y}\:=\:\mathrm{tan}\:\left({x}+\frac{\mathrm{2}\pi}{\mathrm{3}}\right)−\mathrm{tan}\:\left({x}+\frac{\pi}{\mathrm{6}}\right)\:+\mathrm{cos}\:\left({x}+\frac{\pi}{\mathrm{6}}\right) \\ $$$$\mathrm{is}\:\_\_\_ \\ $$

Question Number 96749 Answers: 1 Comments: 0

how we can calclate triple factorial?

$$\mathrm{how}\:\mathrm{we}\:\mathrm{can}\:\mathrm{calclate}\:\mathrm{triple}\:\mathrm{factorial}? \\ $$

Question Number 96748 Answers: 0 Comments: 1

nobody tried question 94184...

$$\mathrm{nobody}\:\mathrm{tried}\:\mathrm{question}\:\mathrm{94184}... \\ $$

Question Number 96730 Answers: 1 Comments: 0

A third of a population has been vaccined against an illness. During the pandemie it is noticed that; out of 15 patients, 2 have been vaccined. Assuming that out of a hundred vaccined, 8 are ill. An individual is choosen at random from this population. Let M imply the individual is ill and V imply the individual has been vaccined. 1\ Determine the probabilities: P(V); P(V/M) and P(M/V) 2\ Calculate P(M∩V) then P(M). Deduce the percentage of patients.

Question Number 96729 Answers: 2 Comments: 1

Prove that ln∣sec(x)+tan(x)∣=tanh^(−1) (sin(x))

$${Prove}\:{that}\:\mathrm{ln}\mid\mathrm{sec}\left({x}\right)+\mathrm{tan}\left({x}\right)\mid=\mathrm{tanh}^{−\mathrm{1}} \left(\mathrm{sin}\left({x}\right)\right) \\ $$

Question Number 96746 Answers: 3 Comments: 2

∫((√x)/((1+x^3 )(√(1−x^3 ))))dx=? ∫((√x)/((1−x^3 )(√(1+x^3 ))))dx=?

$$\int\frac{\sqrt{{x}}}{\left(\mathrm{1}+{x}^{\mathrm{3}} \right)\sqrt{\mathrm{1}−{x}^{\mathrm{3}} }}{dx}=? \\ $$$$\int\frac{\sqrt{{x}}}{\left(\mathrm{1}−{x}^{\mathrm{3}} \right)\sqrt{\mathrm{1}+{x}^{\mathrm{3}} }}{dx}=? \\ $$

Question Number 96715 Answers: 1 Comments: 0

find real solution of equation x^5 +x^4 +1 = 0

$$\mathrm{find}\:\mathrm{real}\:\mathrm{solution}\:\mathrm{of}\:\mathrm{equation} \\ $$$${x}^{\mathrm{5}} +{x}^{\mathrm{4}} +\mathrm{1}\:=\:\mathrm{0} \\ $$

Question Number 96713 Answers: 1 Comments: 0

y^2 (d^2 y/dx^2 )=(dy/dx)

$${y}^{\mathrm{2}} \:\frac{{d}^{\mathrm{2}} {y}}{{dx}^{\mathrm{2}} }=\frac{{dy}}{{dx}} \\ $$