Exercises — Range, max height, time of flight — all derived

KYA poochha gaya hai: hawa mein time $\Rightarrow$ time of flight $T=\frac{2u\sin \theta }{g}$. YE KYUN: "hawa mein" = launch se le kar land karne tak. Yahi $T$ measure karta hai. $R$ nahi (woh distance hai), $H$ nahi (woh height hai). Koi number crunch nahi chahiye — answer hai $T=\frac{2u\sin \theta }{g}$ ko pehchaanna.

Dekho $u$ kaise appear karta hai:

• $R=\frac{u^2\sin 2\theta }{g}$ — $u$ squared hai. $u$ double karne par $2^2=4\times$ range milti hai.
• $T=\frac{2u\sin \theta }{g}$ — $u$ first power par hai. $u$ double karne par $2\times$ time milta hai. YE KYUN MATTER KARTA HAI: kisi variable ka exponent padhna hi physics hai. Range, time se zyada tezi se badhti hai kyunki speed double karne par tum dono zyada der udte ho aur aage tezi se bhi jaate ho — ye do effects milke multiply ho jaate hain.

Time: $T=\frac{2(30)(0.707)}{10}=\frac{42.4}{10}=4.24\text{s}$. Height: $H=\frac{(30{)}^2(0.707{)}^2}{2(10)}=\frac{900(0.5)}{20}=\frac{450}{20}=22.5\text{m}$. $(0.707{)}^2=0.5$ kyun? Kyunki ${\sin }^{2}45^{\circ }=\frac{1}{2}$ exactly hota hai. Range: $R=\frac{(30{)}^2\sin 90^{\circ }}{10}=\frac{900(1)}{10}=90\text{m}$. Sanity check: $45^{\circ }$ par, $R=u^2/g=900/10=90$. ✓ Yeh is speed ke liye maximum possible range hai.

$R=\frac{u^2\sin 2\theta }{g}$ se shuru karo: $20=\frac{(20{)}^2\sin 2\theta }{10}=\frac{400\sin 2\theta }{10}=40\sin 2\theta \Rightarrow \sin 2\theta =\mathrm{0.5.}$ DO answers kyun hain: equation $\sin 2\theta =0.5$ ke valid range $0^{\circ }<2\theta <180^{\circ }$ mein do solutions hain: $2\theta =30^{\circ }\Rightarrow \theta =15^{\circ },2\theta =150^{\circ }\Rightarrow \theta =75^{\circ }.$ Ye complementary hain ($15^{\circ }+75^{\circ }=90^{\circ }$) — low flat shot aur high lob jo ek hi jagah land karte hain. Neeche figure mein twin trajectories dekho.

(a) Ranges. $R_{20}=\frac{625\sin 40^{\circ }}{10}$ aur $R_{70}=\frac{625\sin 140^{\circ }}{10}$. Kyunki $\sin 140^{\circ }=\sin (180^{\circ }-140^{\circ })=\sin 40^{\circ }$, dono $\frac{625(0.643)}{10}\approx 40.2\text{m}$ ke barabar hain. Equal. ✓ Kyun: $20^{\circ }$ aur $70^{\circ }$ complementary hain, aur $\sin 2\theta =\sin (180^{\circ }-2\theta )$ unki ranges identical banata hai.

(b) Heights. $H\propto {\sin }^2\theta$ hai, isliye $\frac{H_{70}}{H_{20}}=\frac{{\sin }^{2}70^{\circ }}{{\sin }^{2}20^{\circ }}={\left(\frac{0.940}{0.342}\right)}^2\approx (2.749{)}^2\approx \mathrm{7.55.}$ Steep wala shot lagbhag 7.5× zyada upar jaata hai jabki usi jagah land karta hai. Equal range ka matlab equal path nahi hota.

Key idea: pehle $u\sin \theta$ nikalo, kyunki $T$ aur $H$ dono sirf vertical component $u_y=u\sin \theta$ par depend karte hain. $T=\frac{2u\sin \theta }{g}$ se: $u\sin \theta =\frac{gT}{2}=\frac{10\cdot 4}{2}=20\text{m/s}$. Height se cross-check: $H=\frac{(u\sin \theta {)}^2}{2g}=\frac{20^2}{2(10)}=\frac{400}{20}=20\text{m}$. ✓ Consistent hai. Toh hum jaante hain $u\sin \theta =20$, lekin yeh ek equation hai do unknowns ($u$ aur $\theta$) ke saath. $T$ aur $H$ akele unhe alag nahi kar sakte — same vertical component wali har launch ke liye same $T$ aur $H$ milega. Conclusion: $u\sin \theta =20\text{m/s}$ fully determined hai, lekin $u$ aur $\theta$ alag-alag nahi — range ya horizontal speed bhi chahiye hoga. Is under-determination ko pehchaanna hi answer hai.

Energy setup. Peak par, sirf vertical part ki velocity zero hoti hai; horizontal speed $u\cos \theta$ bachi rehti hai. Toh climb karne mein jo kinetic energy "kharach" hoti hai woh sirf vertical part ki hai. Unit mass per, launch aur peak ke beech energy conservation: $\frac{1}{2}(u\sin \theta {)}^2=gH\Rightarrow H=\frac{(u\sin \theta {)}^2}{2g}.$ Sirf $u\sin \theta$ kyun? Horizontal speed kabhi nahi badlti aur kabhi height mein convert nahi hoti, isliye dono sides se cancel ho jaati hai. Numbers: $u\sin \theta =20(0.5)=10\text{m/s}$, toh $H=\frac{10^2}{2(10)}=\frac{100}{20}=5\text{m}$. Kinematic formula se match: $H=\frac{(20{)}^2(0.5{)}^2}{2(10)}=\frac{400(0.25)}{20}=5\text{m}$. ✓ Do bilkul alag routes (energy vs. equations of motion) agree karte hain — sahi result ki pehchaan yahi hai.

Do axes mein tod do (projectile motion ka poora point yahi hai — dekho Projectile Motion — components & independence):

• Horizontal velocity constant rehti hai: $v_x=17.32\text{m/s}$.
• Vertical velocity gravity ke under decay karti hai: $v_y=u_y-gt=10-10(1)=0\text{m/s}$. Interpretation: $v_y=0$ at $t=1$s matlab ball usi instant apne peak par hai. (Check: $t_{\text{up}}=u\sin \theta /g=10/10=1$s. ✓) Speed = $\sqrt{v_x^2+v_y^2}=\sqrt{17.32^2+0^2}=17.32\text{m/s}$. Direction: $\tan \varphi =v_y/v_x=0/17.32=0\Rightarrow \varphi =0^{\circ }$ — arc ke top par expected ki tarah bilkul horizontally move kar rahi hai.

Standard $R$ formula kyun fail karta hai: $R=\frac{u^2\sin 2\theta }{g}$ assume karta hai ki launch aur landing same height par hain. Yahan dono $15$m se alag hain, isliye hume vertical equation se re-derive karna hoga. Upar ko positive lo, origin launch point par. Landing $15$m neeche hai, toh $y=-15$. $y=u\sin \theta t-\frac{1}{2}g{t}^2\Rightarrow -15=(20)(0.5)t-\frac{1}{2}(10)t^2=10t-5t^2.$ Standard quadratic form mein rearrange karo: $5t^2-10t-15=0\Rightarrow t^2-2t-3=0\Rightarrow (t-3)(t+1)=0.$ Roots: $t=3\text{s}$ aur $t=-1\text{s}$. Negative root discard karo (launch se pehle); physical flight time hai $\boxed{t=3\text{s}}$. (b) Horizontal distance: horizontal motion uniform hai, isliye $x=u\cos \theta \cdot t=20(0.866)(3)=51.96\approx 52.0\text{m}.$ Gaur karo yeh level-ground range se zyada hai — extra fall time se steady horizontal motion use aage le jaati hai.

$\sin 90^{\circ }=1$, $\cos 90^{\circ }=0$, $\sin (2\cdot 90^{\circ })=\sin 180^{\circ }=0$ ke saath:

• $T=\frac{2(20)(1)}{10}=4\text{s}$ — $2$s upar, $2$s neeche. Sensible hai.
• $H=\frac{(20{)}^2(1{)}^2}{2(10)}=\frac{400}{20}=20\text{m}$ — is speed ke liye highest possible height, kyunki saari speed vertical hai.
• $R=\frac{(20{)}^2\sin 180^{\circ }}{10}=\frac{400(0)}{10}=0\text{m}$. $R=0$ ka interpretation: vertical throw mein koi horizontal component nahi hota ($u\cos 90^{\circ }=0$), toh yeh seedha launch point par wapas aata hai — zero horizontal distance. Formula is degenerate case ko sahi report karta hai.

$\sin 0^{\circ }=0$ ke saath: $T=0$, $H=0$, $R=0$. Teeno zero kyun hain aur yeh sahi kyun hai: formulas assume karte hain ki launch aur landing same height (ground) par hote hain. Agar tum ground se horizontally launch karo, toh vertical component zero hai, isliye gravity ball ko seedha floor mein kheench leti hai — yeh kabhi airborne nahi hoti. $T=H=R=0$ faithfully describe karta hai "ball already ground par hai." L5.1 se contrast: ek cliff se horizontal launch nonzero range dega — lekin phir $T$ ko $y=-h$ se re-derive karna hoga, kyunki same-height assumption toot jaati hai. Yahan ke zeros ek feature hain, bug nahi: ye signal karte hain ki formula ki assumption apni boundary tak pahunch gayi hai.