Question-222123

Question Number 222123 by mr W last updated on 18/Jun/25

Commented by mr W last updated on 18/Jun/25

a ball is released at the point (d, h) and is rebounded from the ground which has the shape of a hyperbola with equation −(x^2 /a^2 )+(((y+b)^2 )/b^2 )=1. if the ball hits exactly the point (0, 0) after the rebound, find h in terms of a, b and d. the collision between the ball and the ground is elastic.

$${a}\:{ball}\:{is}\:{released}\:{at}\:{the}\:{point}\:\left({d},\:{h}\right) \\ $$$${and}\:{is}\:{rebounded}\:{from}\:{the}\:{ground}\: \\ $$$${which}\:{has}\:{the}\:{shape}\:{of}\:{a}\:{hyperbola} \\ $$$${with}\:{equation}\:−\frac{{x}^{\mathrm{2}} }{{a}^{\mathrm{2}} }+\frac{\left({y}+{b}\right)^{\mathrm{2}} }{{b}^{\mathrm{2}} }=\mathrm{1}.\: \\ $$$${if}\:{the}\:{ball}\:{hits}\:{exactly}\:{the}\:{point} \\ $$$$\left(\mathrm{0},\:\mathrm{0}\right)\:{after}\:{the}\:{rebound},\:{find}\:{h}\:{in}\: \\ $$$${terms}\:{of}\:{a},\:{b}\:{and}\:{d}. \\ $$$${the}\:{collision}\:{between}\:{the}\:{ball}\:{and} \\ $$$${the}\:{ground}\:{is}\:{elastic}. \\ $$

Answered by mr W last updated on 18/Jun/25

Commented by mr W last updated on 19/Jun/25

Commented by mr W last updated on 19/Jun/25

y=b((√(1+(x^2 /a^2 )))−1) (dy/dx)=((bx)/(a^2 (√(1+(x^2 /a^2 ))))) at x=d: y_P =b((√(1+(d^2 /a^2 )))−1) tan θ=((bd)/(a^2 (√(1+(d^2 /a^2 ))))) v=(√(2g(h−y_P ))) t=(d/(v sin 2θ)) −y_P =v cos 2θ t−((gt^2 )/2) −b((√(1+(d^2 /a^2 )))−1)=(d/(tan 2θ))−(d^2 /(4(h−y_P )))(1+(1/(tan^2 2θ))) ⇒h=b((√(1+(d^2 /a^2 )))−1)+((d^2 (1+(1/(tan^2 2θ))))/(4b((√(1+(d^2 /a^2 )))+(d/(b tan 2θ))−1)))

$${y}={b}\left(\sqrt{\mathrm{1}+\frac{{x}^{\mathrm{2}} }{{a}^{\mathrm{2}} }}−\mathrm{1}\right) \\ $$$$\frac{{dy}}{{dx}}=\frac{{bx}}{{a}^{\mathrm{2}} \sqrt{\mathrm{1}+\frac{{x}^{\mathrm{2}} }{{a}^{\mathrm{2}} }}} \\ $$$${at}\:{x}={d}: \\ $$$${y}_{{P}} ={b}\left(\sqrt{\mathrm{1}+\frac{{d}^{\mathrm{2}} }{{a}^{\mathrm{2}} }}−\mathrm{1}\right) \\ $$$$\mathrm{tan}\:\theta=\frac{{bd}}{{a}^{\mathrm{2}} \sqrt{\mathrm{1}+\frac{{d}^{\mathrm{2}} }{{a}^{\mathrm{2}} }}} \\ $$$${v}=\sqrt{\mathrm{2}{g}\left({h}−{y}_{{P}} \right)} \\ $$$${t}=\frac{{d}}{{v}\:\mathrm{sin}\:\mathrm{2}\theta} \\ $$$$−{y}_{{P}} ={v}\:\mathrm{cos}\:\mathrm{2}\theta\:{t}−\frac{{gt}^{\mathrm{2}} }{\mathrm{2}} \\ $$$$−{b}\left(\sqrt{\mathrm{1}+\frac{{d}^{\mathrm{2}} }{{a}^{\mathrm{2}} }}−\mathrm{1}\right)=\frac{{d}}{\mathrm{tan}\:\mathrm{2}\theta}−\frac{{d}^{\mathrm{2}} }{\mathrm{4}\left({h}−{y}_{{P}} \right)}\left(\mathrm{1}+\frac{\mathrm{1}}{\mathrm{tan}^{\mathrm{2}} \:\mathrm{2}\theta}\right) \\ $$$$\Rightarrow{h}={b}\left(\sqrt{\mathrm{1}+\frac{{d}^{\mathrm{2}} }{{a}^{\mathrm{2}} }}−\mathrm{1}\right)+\frac{{d}^{\mathrm{2}} \left(\mathrm{1}+\frac{\mathrm{1}}{\mathrm{tan}^{\mathrm{2}} \:\mathrm{2}\theta}\right)}{\mathrm{4}{b}\left(\sqrt{\mathrm{1}+\frac{{d}^{\mathrm{2}} }{{a}^{\mathrm{2}} }}+\frac{{d}}{{b}\:\mathrm{tan}\:\mathrm{2}\theta}−\mathrm{1}\right)} \\ $$

Commented by fantastic last updated on 19/Jun/25

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