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Question Number 89319 by Rio Michael last updated on 16/Apr/20
A ball is projected from a point O with an initial velocity u and angle θ with the horizontal ground. Given that it travels such that it just clears two walls of height h and distances 2h and 4h from O respectively. (a) find the tangent of the angle θ (b) The time of flight of the ball (c) The range of the ball.
$$\:\mathrm{A}\:\mathrm{ball}\:\mathrm{is}\:\mathrm{projected}\:\mathrm{from}\:\mathrm{a}\:\mathrm{point}\:\mathrm{O}\:\mathrm{with}\:\mathrm{an}\:\mathrm{initial} \\ $$$$\mathrm{velocity}\:{u}\:\mathrm{and}\:\mathrm{angle}\:\theta\:\mathrm{with}\:\mathrm{the}\:\mathrm{horizontal}\:\mathrm{ground}. \\ $$$$\:\mathrm{Given}\:\mathrm{that}\:\mathrm{it}\:\mathrm{travels}\:\mathrm{such}\:\mathrm{that}\:\mathrm{it}\:\mathrm{just}\:\mathrm{clears}\:\mathrm{two}\:\mathrm{walls} \\ $$$$\mathrm{of}\:\mathrm{height}\:{h}\:\mathrm{and}\:\mathrm{distances}\:\mathrm{2}{h}\:\mathrm{and}\:\mathrm{4}{h}\:\mathrm{from}\:\mathrm{O}\:\mathrm{respectively}. \\ $$$$\left(\mathrm{a}\right)\:\mathrm{find}\:\mathrm{the}\:\mathrm{tangent}\:\mathrm{of}\:\mathrm{the}\:\mathrm{angle}\:\theta \\ $$$$\:\left(\mathrm{b}\right)\:\mathrm{The}\:\mathrm{time}\:\mathrm{of}\:\mathrm{flight}\:\mathrm{of}\:\mathrm{the}\:\mathrm{ball} \\ $$$$\left(\mathrm{c}\right)\:\mathrm{The}\:\mathrm{range}\:\mathrm{of}\:\mathrm{the}\:\mathrm{ball}. \\ $$
Answered by mr W last updated on 17/Apr/20
since the track of the ball is a parabola, you can get the answer without much calcuation: (c) due to symmetry, the range is L=2h+2h+2h=6h the maximum height of the ball h_(max) (h_(max) /(h_(max) −h))=(((3h)/h))^2 =9 ⇒h_(max) =(9/8)h (a) tan θ=((2h_(max) )/(3h))=(3/4) (b) total flight time =T h_(max) =(1/2)g((T/2))^2 =((9h)/8) ⇒T=3(√(h/(2g)))
$${since}\:{the}\:{track}\:{of}\:{the}\:{ball}\:{is}\:{a}\:{parabola}, \\ $$$${you}\:{can}\:{get}\:{the}\:{answer}\:{without}\:{much} \\ $$$${calcuation}: \\ $$$$\left({c}\right)\:{due}\:{to}\:{symmetry},\:{the}\:{range}\:{is} \\ $$$${L}=\mathrm{2}{h}+\mathrm{2}{h}+\mathrm{2}{h}=\mathrm{6}{h} \\ $$$$ \\ $$$${the}\:{maximum}\:{height}\:{of}\:{the}\:{ball}\:{h}_{{max}} \\ $$$$\frac{{h}_{{max}} }{{h}_{{max}} −{h}}=\left(\frac{\mathrm{3}{h}}{{h}}\right)^{\mathrm{2}} =\mathrm{9} \\ $$$$\Rightarrow{h}_{{max}} =\frac{\mathrm{9}}{\mathrm{8}}{h} \\ $$$$ \\ $$$$\left({a}\right) \\ $$$$\mathrm{tan}\:\theta=\frac{\mathrm{2}{h}_{{max}} }{\mathrm{3}{h}}=\frac{\mathrm{3}}{\mathrm{4}} \\ $$$$ \\ $$$$\left({b}\right) \\ $$$${total}\:{flight}\:{time}\:={T} \\ $$$${h}_{{max}} =\frac{\mathrm{1}}{\mathrm{2}}{g}\left(\frac{{T}}{\mathrm{2}}\right)^{\mathrm{2}} =\frac{\mathrm{9}{h}}{\mathrm{8}} \\ $$$$\Rightarrow{T}=\mathrm{3}\sqrt{\frac{{h}}{\mathrm{2}{g}}} \\ $$
Commented by peter frank last updated on 17/Apr/20
sorry sir where 3h came from?
$${sorry}\:{sir}\:{where}\:\mathrm{3}{h}\:{came}\:{from}? \\ $$
Commented by mr W last updated on 17/Apr/20
horizontal distance of the ball to O is 3h when it is at its highest position.
$${horizontal}\:{distance}\:{of}\:{the}\:{ball}\:{to}\:{O} \\ $$$${is}\:\mathrm{3}{h}\:{when}\:{it}\:{is}\:{at}\:{its}\:{highest}\:{position}. \\ $$
Commented by mr W last updated on 17/Apr/20
Commented by peter frank last updated on 17/Apr/20
thank you now its clear.
$${thank}\:{you}\:{now}\:{its}\:{clear}. \\ $$
Commented by Rio Michael last updated on 17/Apr/20
sir why would you say it travels a distance of 2h after the second wall?.
$$\mathrm{sir}\:\mathrm{why}\:\mathrm{would}\:\mathrm{you}\:\mathrm{say}\:\mathrm{it}\:\mathrm{travels}\:\mathrm{a}\:\mathrm{distance}\:\mathrm{of}\:\mathrm{2}{h}\:\mathrm{after}\:\mathrm{the}\:\mathrm{second}\:\mathrm{wall}?. \\ $$
Commented by mr W last updated on 17/Apr/20
at x=2h the ball has a height h at x=4h the ball has a height h ⇒in the middle, i.e. at x=3h, the ball reaches h_(max) , because the parabola is symmetric right or left to the highest point. you can get the rest.
$${at}\:{x}=\mathrm{2}{h}\:{the}\:{ball}\:{has}\:{a}\:{height}\:{h} \\ $$$${at}\:{x}=\mathrm{4}{h}\:{the}\:{ball}\:{has}\:{a}\:{height}\:{h} \\ $$$$\Rightarrow{in}\:{the}\:{middle},\:{i}.{e}.\:{at}\:{x}=\mathrm{3}{h},\:{the} \\ $$$${ball}\:{reaches}\:{h}_{{max}} ,\:{because}\:{the}\:{parabola} \\ $$$${is}\:{symmetric}\:{right}\:{or}\:{left}\:{to}\:{the} \\ $$$${highest}\:{point}.\:{you}\:{can}\:{get}\:{the}\:{rest}. \\ $$
Commented by Rio Michael last updated on 17/Apr/20
thank you sir
$$\mathrm{thank}\:\mathrm{you}\:\mathrm{sir} \\ $$
Commented by peter frank last updated on 20/Apr/20
why you square (((3h)/h))^2 ?
$${why}\:{you}\:{square}\:\left(\frac{\mathrm{3}{h}}{{h}}\right)^{\mathrm{2}} ? \\ $$
Commented by peter frank last updated on 20/Apr/20
where is square came from?
$${where}\:{is}\:{square}\:{came} \\ $$$${from}? \\ $$
Commented by mr W last updated on 21/Apr/20
the track of ball is a parabola! with a parabola e.g. y=ax^2 we have y_1 =ax_1 ^2 y_2 =ax_2 ^2 (y_1 /y_2 )=((x_1 /x_2 ))^2
$${the}\:{track}\:{of}\:{ball}\:{is}\:{a}\:{parabola}! \\ $$$${with}\:{a}\:{parabola}\:{e}.{g}.\:{y}={ax}^{\mathrm{2}} \:{we}\:{have} \\ $$$${y}_{\mathrm{1}} ={ax}_{\mathrm{1}} ^{\mathrm{2}} \\ $$$${y}_{\mathrm{2}} ={ax}_{\mathrm{2}} ^{\mathrm{2}} \\ $$$$\frac{{y}_{\mathrm{1}} }{{y}_{\mathrm{2}} }=\left(\frac{{x}_{\mathrm{1}} }{{x}_{\mathrm{2}} }\right)^{\mathrm{2}} \\ $$
Commented by peter frank last updated on 21/Apr/20
thank you
$${thank}\:{you} \\ $$

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