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a particle is projected from a point at a height 3h metres above a horizontal play ground. the direction of the projectile makes an angle α with the horizontal through the point of projection. show that if th greatest height reached above the point lc projection is h metres, then the horizontal distance travelled by the particle before striking the plane is 6h cotα metres. Find the vertical and horizontal component of the speed of the particle just before it hits the ground.

Question Number 91931 Answers: 0 Comments: 3

∫_0 ^1 ln(Γ(x)) dx

$$\int_{\mathrm{0}} ^{\mathrm{1}} {ln}\left(\Gamma\left({x}\right)\right)\:{dx} \\ $$

Question Number 91930 Answers: 0 Comments: 2

lim_(x→0) cos (1/x)=

$$\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}cos}\:\frac{\mathrm{1}}{{x}}= \\ $$

Question Number 91785 Answers: 0 Comments: 5

given these bellow sequenses . find fifth term 1). (1/5), (1/4) , (3/(11)), (2/7),... 2). (3/5), ((−9)/(11)) , ((−25)/(19)), ((57)/(35)),... 3). (1/6), (7/(11)) , ((13)/(16)), ((19)/(21)),... 4). 4,− (2/3) , −(4/(13)),− (1/5),... 5). 2, 9 , 28 , 65,...

$$\mathrm{given}\:\:\mathrm{these}\:\mathrm{bellow}\:\mathrm{sequenses}\:.\: \\ $$$$\mathrm{find}\:\mathrm{fifth}\:\mathrm{term} \\ $$$$ \\ $$$$\left.\mathrm{1}\right).\:\:\frac{\mathrm{1}}{\mathrm{5}},\:\frac{\mathrm{1}}{\mathrm{4}}\:,\:\frac{\mathrm{3}}{\mathrm{11}},\:\frac{\mathrm{2}}{\mathrm{7}},... \\ $$$$\left.\mathrm{2}\right).\:\:\frac{\mathrm{3}}{\mathrm{5}},\:\frac{−\mathrm{9}}{\mathrm{11}}\:,\:\frac{−\mathrm{25}}{\mathrm{19}},\:\frac{\mathrm{57}}{\mathrm{35}},... \\ $$$$\left.\mathrm{3}\right).\:\:\:\frac{\mathrm{1}}{\mathrm{6}},\:\frac{\mathrm{7}}{\mathrm{11}}\:,\:\frac{\mathrm{13}}{\mathrm{16}},\:\frac{\mathrm{19}}{\mathrm{21}},... \\ $$$$\left.\mathrm{4}\right).\:\:\:\mathrm{4},−\:\frac{\mathrm{2}}{\mathrm{3}}\:,\:−\frac{\mathrm{4}}{\mathrm{13}},−\:\frac{\mathrm{1}}{\mathrm{5}},... \\ $$$$\left.\mathrm{5}\right).\:\:\mathrm{2},\:\mathrm{9}\:,\:\mathrm{28}\:,\:\mathrm{65},... \\ $$

Question Number 91783 Answers: 0 Comments: 1

1). (1/2), (2/3), (3/4) , (4/5), ..., .... 2). 4,6,10,18,34,...,.... 3). 5,7,11,19,35,...,.... 4). 4,6,10,18,34,...,.... 5). 4,11,30,85,248,...,...

$$\left.\mathrm{1}\right).\:\:\frac{\mathrm{1}}{\mathrm{2}},\:\frac{\mathrm{2}}{\mathrm{3}},\:\frac{\mathrm{3}}{\mathrm{4}}\:,\:\frac{\mathrm{4}}{\mathrm{5}},\:...,\:.... \\ $$$$\left.\mathrm{2}\right).\:\:\mathrm{4},\mathrm{6},\mathrm{10},\mathrm{18},\mathrm{34},...,.... \\ $$$$\left.\mathrm{3}\right).\:\:\mathrm{5},\mathrm{7},\mathrm{11},\mathrm{19},\mathrm{35},...,.... \\ $$$$\left.\mathrm{4}\right).\:\:\mathrm{4},\mathrm{6},\mathrm{10},\mathrm{18},\mathrm{34},...,.... \\ $$$$\left.\mathrm{5}\right).\:\:\mathrm{4},\mathrm{11},\mathrm{30},\mathrm{85},\mathrm{248},...,... \\ $$

Question Number 91715 Answers: 0 Comments: 4

Question Number 91655 Answers: 2 Comments: 12

A particle is projected with an intial velocity of u ms^(−1) at an angle α to the ground from a point O on the ground. Given that it clears two walls of hieght h and distances 2h and 4h respectively from O. (a) find the tangent of α (b) the maximum hieght (c) the range and period of the particle (d) show that u^2 = (4/(26)) gh please sir can you help me using the actual equations of projectile motion?

$$\mathrm{A}\:\mathrm{particle}\:\mathrm{is}\:\mathrm{projected}\:\mathrm{with}\:\mathrm{an}\:\mathrm{intial}\:\mathrm{velocity}\:\mathrm{of}\:{u}\:\mathrm{ms}^{−\mathrm{1}} \:\mathrm{at}\:\mathrm{an}\:\mathrm{angle}\:\alpha\:\mathrm{to}\: \\ $$$$\mathrm{the}\:\mathrm{ground}\:\mathrm{from}\:\mathrm{a}\:\mathrm{point}\:\mathrm{O}\:\mathrm{on}\:\mathrm{the}\:\mathrm{ground}.\:\mathrm{Given}\:\mathrm{that}\:\mathrm{it}\:\mathrm{clears} \\ $$$$\mathrm{two}\:\mathrm{walls}\:\mathrm{of}\:\mathrm{hieght}\:{h}\:\mathrm{and}\:\mathrm{distances}\:\mathrm{2h}\:\mathrm{and}\:\mathrm{4h}\:\mathrm{respectively}\:\mathrm{from}\:\mathrm{O}. \\ $$$$\left(\mathrm{a}\right)\:\mathrm{find}\:\mathrm{the}\:\mathrm{tangent}\:\mathrm{of}\:\alpha \\ $$$$\left(\mathrm{b}\right)\:\mathrm{the}\:\mathrm{maximum}\:\mathrm{hieght} \\ $$$$\left(\mathrm{c}\right)\:\mathrm{the}\:\mathrm{range}\:\mathrm{and}\:\mathrm{period}\:\mathrm{of}\:\mathrm{the}\:\mathrm{particle} \\ $$$$\left(\mathrm{d}\right)\:\mathrm{show}\:\mathrm{that}\:{u}^{\mathrm{2}} \:=\:\frac{\mathrm{4}}{\mathrm{26}}\:\mathrm{g}{h}\: \\ $$$$\mathrm{please}\:\mathrm{sir}\:\mathrm{can}\:\mathrm{you}\:\mathrm{help}\:\mathrm{me}\:\mathrm{using}\:\mathrm{the}\:\mathrm{actual}\:\mathrm{equations}\:\mathrm{of}\:\mathrm{projectile}\:\mathrm{motion}? \\ $$$$ \\ $$

Question Number 91654 Answers: 1 Comments: 0

th position vector of a particle p of mass 3 kg is given by r = [(cos 2t) i + (sin 2t)j] m given that p was intitialy at rest. find the cartesian equation of its path and describe it.

$$\mathrm{th}\:\mathrm{position}\:\mathrm{vector}\:\mathrm{of}\:\mathrm{a}\:\mathrm{particle}\:{p}\:\mathrm{of}\:\mathrm{mass}\:\mathrm{3}\:\mathrm{kg}\:\mathrm{is}\:\mathrm{given}\:\mathrm{by}\: \\ $$$$\:{r}\:=\:\left[\left(\mathrm{cos}\:\mathrm{2}{t}\right)\:{i}\:+\:\left(\mathrm{sin}\:\mathrm{2}{t}\right){j}\right]\:\mathrm{m} \\ $$$$\mathrm{given}\:\mathrm{that}\:{p}\:\:\mathrm{was}\:\mathrm{intitialy}\:\mathrm{at}\:\mathrm{rest}. \\ $$$$\mathrm{find}\:\mathrm{the}\:\mathrm{cartesian}\:\mathrm{equation}\:\mathrm{of}\:\mathrm{its}\:\mathrm{path}\:\mathrm{and}\:\mathrm{describe}\:\mathrm{it}. \\ $$

Question Number 91638 Answers: 1 Comments: 0

Find the general term: Σ_(n=1) ^k n^n

$${Find}\:{the}\:{general}\:{term}: \\ $$$$\underset{{n}=\mathrm{1}} {\overset{{k}} {\sum}}{n}^{{n}} \\ $$

Question Number 91613 Answers: 1 Comments: 4

solve without using l′hopital lim_(x→e) ((ln(x)−1)/((e/x)−1))

$${solve}\:{without}\:{using}\:{l}'{hopital} \\ $$$$\underset{{x}\rightarrow{e}} {{lim}}\frac{{ln}\left({x}\right)−\mathrm{1}}{\frac{{e}}{{x}}−\mathrm{1}} \\ $$

Question Number 91604 Answers: 0 Comments: 0

Question Number 91588 Answers: 0 Comments: 2

what is f^(−1) for f(x)=⌊x⌋??

$${what}\:{is}\:{f}^{−\mathrm{1}} \:{for}\:{f}\left({x}\right)=\lfloor{x}\rfloor?? \\ $$

Question Number 91476 Answers: 0 Comments: 2

Find the slope of the tangent line to the graph of: y^4 +3y−4x^3 =5x+1 at the point P (1, −2)

$$\mathrm{Find}\:\mathrm{the}\:\mathrm{slope}\:\mathrm{of}\:\mathrm{the}\:\mathrm{tangent}\: \\ $$$$\mathrm{line}\:\mathrm{to}\:\mathrm{the}\:\mathrm{graph}\:\mathrm{of}: \\ $$$$\mathrm{y}^{\mathrm{4}} +\mathrm{3y}−\mathrm{4x}^{\mathrm{3}} =\mathrm{5x}+\mathrm{1}\:\:\mathrm{at}\:\mathrm{the}\:\mathrm{point} \\ $$$$\mathrm{P}\:\left(\mathrm{1},\:−\mathrm{2}\right) \\ $$

Question Number 91421 Answers: 2 Comments: 2

∫_1 ^3 (1/(x(√(3x^2 +2x−1))))dx

$$\int_{\mathrm{1}} ^{\mathrm{3}} \frac{\mathrm{1}}{{x}\sqrt{\mathrm{3}{x}^{\mathrm{2}} +\mathrm{2}{x}−\mathrm{1}}}{dx} \\ $$

Question Number 91375 Answers: 0 Comments: 3

x4+2x3−5x2+6x+2/x2−2x+2^

$${x}\mathrm{4}+\mathrm{2}{x}\mathrm{3}−\mathrm{5}{x}\mathrm{2}+\mathrm{6}{x}+\mathrm{2}/{x}\mathrm{2}−\mathrm{2}{x}+\mathrm{2}^{} \\ $$

Question Number 91374 Answers: 0 Comments: 0

A car of mass 700 kg has maximum power P ,at all times, there is a non gravitational R to the motion of the car. the car moves along an inclined of angle θ where 10 sinθ = 1. The maximum speed of the car up the plane is is half the value of the speed down the plane. (a) find the value of R. on level road the car has speed of 20 ms^(−1) . (b) find the value of P.

$$\mathrm{A}\:\mathrm{car}\:\mathrm{of}\:\mathrm{mass}\:\mathrm{700}\:\mathrm{kg}\:\mathrm{has}\:\mathrm{maximum}\:\mathrm{power}\:{P}\:\:,\mathrm{at}\:\mathrm{all}\:\mathrm{times}, \\ $$$$\mathrm{there}\:\mathrm{is}\:\mathrm{a}\:\mathrm{non}\:\mathrm{gravitational}\:{R}\:\mathrm{to}\:\mathrm{the}\:\mathrm{motion}\:\mathrm{of}\:\mathrm{the}\:\mathrm{car}. \\ $$$$\mathrm{the}\:\mathrm{car}\:\mathrm{moves}\:\mathrm{along}\:\mathrm{an}\:\mathrm{inclined}\:\mathrm{of}\:\mathrm{angle}\:\theta\:\mathrm{where}\:\mathrm{10}\:\mathrm{sin}\theta\:=\:\mathrm{1}.\:\mathrm{The} \\ $$$$\mathrm{maximum}\:\mathrm{speed}\:\mathrm{of}\:\mathrm{the}\:\mathrm{car}\:\mathrm{up}\:\mathrm{the}\:\mathrm{plane}\:\mathrm{is}\:\mathrm{is}\:\mathrm{half}\:\mathrm{the}\:\mathrm{value}\:\mathrm{of}\:\mathrm{the} \\ $$$$\mathrm{speed}\:\mathrm{down}\:\mathrm{the}\:\mathrm{plane}. \\ $$$$\left(\mathrm{a}\right)\:\mathrm{find}\:\mathrm{the}\:\mathrm{value}\:\mathrm{of}\:{R}. \\ $$$$\:\mathrm{on}\:\mathrm{level}\:\mathrm{road}\:\mathrm{the}\:\mathrm{car}\:\mathrm{has}\:\mathrm{speed}\:\mathrm{of}\:\mathrm{20}\:\mathrm{ms}^{−\mathrm{1}} . \\ $$$$\left(\mathrm{b}\right)\:\mathrm{find}\:\mathrm{the}\:\mathrm{value}\:\mathrm{of}\:{P}. \\ $$

Question Number 91303 Answers: 0 Comments: 3

Question Number 91302 Answers: 2 Comments: 0

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