Range, max height, time of flight — all derived: Worked examples

1.1.20 · D3 · Physics › Measurement, Vectors & Kinematics › Range, max height, time of flight — all derived

Intuition Yeh page kis kaam ki hai

Parent note ne tumhe teen boxed formulas diye. Formulas tab tak bekar hain jab tak tum unhe har possible situation mein throw karke nahi dekh lete ki woh behave kaise karte hain. Yeh page ek shooting gallery hai: hum har tarah ke projectile problems list karte hain jo pooche ja sakte hain, phir har ek ko ek fully worked example se knock down karte hain. Jab tum finish karo, koi bhi exam question tumhe aisa scenario nahi dikha sakta jo tum pehle se dekh nahi chuke.

Hum sirf teen tools reuse karte hain, sab parent mein banaye hue: $T = \frac{2 u s i n θ}{g}, H = \frac{u ^{2} s i n ^{2} θ}{2 g}, R = \frac{u ^{2} s i n 2 θ}{g}$ Yahan $u$ launch ki speed hai (metres per second mein), $θ$ launch angle hai jo flat ground se upar measure kiya gaya hai, $g$ gravity ki strength hai ( $10 m/s^{2}$ jab tak aur na bataya jaye), $T$ hawa mein time hai (seconds mein), $H$ top height hai (metres mein), $R$ flat distance travelled hai (metres mein).

Scenario matrix

Har projectile question actually neeche diye gaye cells mein se ek hota hai. Right column us example ka naam batata hai jo use clear karta hai.

#	Case class	Kya tricky hai	Cleared by
A	Standard slant ( $0 < θ < 9 0^{\circ}$ )	seedha "plug in" wala case	Example 1
B	Complementary pair ( $θ$ aur $9 0^{\circ} - θ$ )	same range, alag path	Example 2
C	Degenerate: $θ = 0^{\circ}$	seedha flat ground par fire kiya	Example 3
D	Degenerate: $θ = 9 0^{\circ}$	seedha upar, $R = 0$	Example 3
E	Given $T$ ya $H$ , find $u$ / $θ$	formula ko ulta chalao	Example 4
F	Unequal heights (cliff se)	ground formula banned, re-derive karo	Example 5
G	Real-world word problem	English → symbols mein translate karo	Example 6
H	Limiting behaviour ( $u \to$ bada, $g \to$ chhota)	answers scale kaise karte hain	Example 7
I	Exam twist: do facts, hidden $θ$	$H$ aur $R$ combine karo	Example 8

Har cell A–I neeche kam se kam ek baar appear karta hai.

Example 1 — Cell A: standard slant

Worked example Statement

Ek pathar $u = 25 m/s$ , $θ = 3 7^{\circ}$ , $g = 10 m/s^{2}$ par launch kiya gaya hai. $T$ , $H$ , $R$ nikalo. $sin 3 7^{\circ} = 0.6$ , $cos 3 7^{\circ} = 0.8$ use karo.

Forecast: aage padhne se pehle guess karo: kya $H$ , $R$ se zyada hoga ya kam? (Hint: $R$ mein $sin 2 θ$ hai, jo $sin 7 4^{\circ} \approx 0.96$ ke paas hai — almost maximum — to $R$ , $H$ ko bahut peeche chhod dega.)

Step 1 — Time of flight. $T = \frac{2 u sin θ}{g}$ use karo. $T = \frac{2 ( 25 ) ( 0.6 )}{10} = \frac{30}{10} = 3 s$ Yeh step kyun? Vertical motion akela clock set karta hai; $sin θ$ launch speed ka upward hissa pick out karta hai.

Step 2 — Max height. $H = \frac{u ^{2} sin ^{2} θ}{2 g}$ use karo. $H = \frac{( 25 ) ^{2} ( 0.6 ) ^{2}}{2 ( 10 )} = \frac{625 \cdot 0.36}{20} = \frac{225}{20} = 11.25 m$ Yeh step kyun? $sin θ$ ko square karna AUR $u$ ko square karna — height "energy-like" hai, speed ke saath fast badhti hai.

Step 3 — Range. $R = \frac{u ^{2} sin 2 θ}{g}$ use karo, $2 θ = 7 4^{\circ}$ , $sin 7 4^{\circ} \approx 0.9613$ . $R = \frac{625 ( 0.9613 )}{10} \approx 60.1 m$ Yeh step kyun? Range = steady forward speed $\times$ time aloft; identity $2 sin θ cos θ = sin 2 θ$ dono ko ek clean term mein fold kar deti hai.

Verify: dimensions — $T$ hai $\frac{m/s}{m/s ^{2}} = s$ ✓; $H, R$ hain $\frac{( m/s ) ^{2}}{m/s ^{2}} = m$ ✓. $R$ ko long way se cross-check karo: $u cos θ \cdot T = 25 (0.8) (3) = 60 m$ , jo $60.1$ se match karta hai (rounding) ✓. Aur $R ≫ H$ jaise forecast kiya tha ✓.

Example 2 — Cell B: complementary twin

Worked example Statement

Wahi pathar fire karo ( $u = 25$ , $g = 10$ ) lekin $θ = 5 3^{\circ}$ par. Dikhao ki uska range Example 1 ke barabar hai, phir bhi uski flight zyada tall aur zyada time lene wali hai. ( $sin 5 3^{\circ} = 0.8, cos 5 3^{\circ} = 0.6$ .)

Forecast: $5 3^{\circ} = 9 0^{\circ} - 3 7^{\circ}$ . Guess karo: $T, H, R$ mein se kaunsa same rahega?

Step 1 — Range. $2 θ = 10 6^{\circ}$ , aur $sin 10 6^{\circ} = sin (18 0^{\circ} - 10 6^{\circ}) = sin 7 4^{\circ} \approx 0.9613$ . $R = \frac{625 ( 0.9613 )}{10} \approx 60.1 m$ Yeh step kyun? $sin$ $9 0^{\circ}$ ke baare mein mirror-symmetric hai, isliye $2 θ$ aur $18 0^{\circ} - 2 θ$ same range dete hain — yahi wajah hai ki complementary angles twins hain.

Step 2 — Height. $H = \frac{625 ( 0.8 ) ^{2}}{20} = \frac{625 \cdot 0.64}{20} = 20 m$ . Yeh step kyun? Ab $sin θ$ bada hai (0.8 vs 0.6), to upward launch slice badi hai — zyada upar chadtha hai.

Step 3 — Time. $T = \frac{2 ( 25 ) ( 0.8 )}{10} = 4 s$ . Yeh step kyun? Zyada upward speed ⇒ rise aur fall mein zyada waqt.

Verify: $R$ Example 1 se identical hai (60.1 m) ✓ — the twin. Lekin $H$ 11.25 m se badh kar 20 m ho gaya aur $T$ 3 s se 4 s: same landing spot, ek taller, slower arc. Sanity check: $u cos θ \cdot T = 25 (0.6) (4) = 60$ m ✓.

Neeche dono arcs dekho. Figure mein dono throws ek hi board par draw hain: chalk-blue curve $3 7^{\circ}$ wala flat-fast shot hai, chalk-pink curve $5 3^{\circ}$ wala tall-slow shot hai. Apni ungali se dono trace karo aur notice karo ki woh ground par same pale-yellow landing dot par milte hain — range genuinely identical hai — jabki pink wala almost do guna upar peak karta hai. Woh ek picture hi complementary-twin idea hai.

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

Blue arc apna peak $11.25$ m par jaldi reach kar leta hai aur jaldi neeche aata hai; pink arc $20$ m tak jaata hai aur zyada time leta hai, phir bhi dono $60.1$ m par land karte hain. Same landing spot, bilkul alag journey — exactly yahi matlab hai "same range, different path" ka.

Example 3 — Cells C & D: do degenerate throws

Worked example Statement

$u = 25$ , $g = 10$ ke saath, (i) $θ = 0^{\circ}$ (dead flat) aur (ii) $θ = 9 0^{\circ}$ (seedha upar) par sab kuch evaluate karo.

Forecast: har case mein kaunsi quantities zero ho jaayengi? Pehle motion picture karo.

Case (i), $θ = 0^{\circ}$ : $sin 0 = 0$ , $cos 0 = 1$ , $sin (2 \cdot 0) = sin 0 = 0$ . $T = \frac{2 u ( 0 )}{g} = 0, H = \frac{u ^{2} ( 0 )}{2 g} = 0, R = \frac{u ^{2} ( 0 )}{g} = 0$ Yeh step kyun? Ground level se perfectly horizontally launch kiya hua ball ka upward speed zero hai, isliye woh already "land" kar chuka hai — koi air time nahi, koi height nahi, koi travel nahi. (Use table se launch karo to Example 5 ka cliff case milta hai.)

Case (ii), $θ = 9 0^{\circ}$ : $sin 9 0^{\circ} = 1$ , $cos 9 0^{\circ} = 0$ , $sin 18 0^{\circ} = 0$ . $T = \frac{2 u ( 1 )}{g} = \frac{2 ( 25 )}{10} = 5 s, H = \frac{u ^{2} ( 1 )}{2 g} = \frac{625}{20} = 31.25 m, R = \frac{u ^{2} ( 0 )}{g} = 0$ Yeh step kyun? Seedha upar matlab poori speed climbing mein jaati hai (max height) aur kuch bhi forward nahi (zero range). Woh seedha wapas wahan aati hai jahan se gayi — $R = 0$ sahi hai, koi galti nahi.

Verify: $θ = 9 0^{\circ}$ sabse bada possible $H$ deta hai (kyunki $sin^{2} θ$ 1 par max hota hai) phir bhi zero range — yeh perfect illustration hai ki height aur range trade off hote hain ✓. Units sab consistent hain ✓.

Example 4 — Cell E: formula ulta chalana

Worked example Statement

$θ = 3 0^{\circ}$ par launch kiya gaya projectile exactly $T = 2 s$ tak hawa mein rehta hai ( $g = 10$ ). Launch speed $u$ nikalo, phir uska range.

Forecast: hume output $T$ diya gaya hai aur input $u$ recover karna hai. Guess karo kya $u$ , $T \times g = 20$ se bada hoga ya chhota.

Step 1 — Time of flight invert karo. $T = \frac{2 u sin θ}{g}$ se shuru karo aur $u$ ke liye solve karo: $u = \frac{g T}{2 s i n θ} = \frac{( 10 ) ( 2 )}{2 ( 0.5 )} = \frac{20}{1} = 20 m/s$ Yeh step kyun? Algebra reversible hai; wahi formula jo $u$ se $T$ predict karta hai, woh $T$ known hone par $u$ ke liye solve bhi karne deta hai.

Step 2 — Ab range. $R = \frac{u ^{2} sin 2 θ}{g} = \frac{400 sin 6 0 ^{\circ}}{10} = \frac{400 ( 0.8660 )}{10} \approx 34.64 m$ . Yeh step kyun? Ek baar $u$ pata chal jaaye to projectile fully determined ho jaata hai; baaki har quantity ussi se nikalti hai.

Verify: $u = 20$ ko forward feed karo: $T = \frac{2 ( 20 ) ( 0.5 )}{10} = 2$ s ✓ — diye gaye time se match karta hai. Yeh wahi ball hai jaise parent note ke Example 1 mein, to $R \approx 34.6$ m ✓.

Example 5 — Cell F: cliff se (formula forbidden!)

Worked example Statement

Ek ball $20 m$ oonchi cliff ke upar se $u = 15 m/s$ , $θ = 5 3^{\circ}$ above horizontal par throw ki jaati hai ( $g = 10$ , $sin 5 3^{\circ} = 0.8$ , $cos 5 3^{\circ} = 0.6$ ). Neeche ground se takraane ka total time aur base se horizontal distance nikalo.

Forecast: boxed $R = \frac{u ^{2} s i n 2 θ}{g}$ assume karta hai ki woh launch height par land karta hai. Yahan woh neeche land karta hai. Guess karo: kya asli time flat-ground $T$ se zyada hoga ya kam?

Step 1 — Full vertical equation set up karo. Upar positive lo, launch point origin. Landing $20$ m neeche hai, to displacement $y = - 20$ . $y = u sin θ t - \frac{1}{2} g t^{2} \Rightarrow - 20 = 15 (0.8) t - 5 t^{2} = 12 t - 5 t^{2}$ Yeh step kyun? Ground formula banned hai kyunki launch aur landing heights alag hain; hume raw quadratic $y \neq = 0$ ke saath solve karna hoga.

Step 2 — Quadratic $5 t^{2} - 12 t - 20 = 0$ solve karo. $t = \frac{12 \pm 144 + 400}{10} = \frac{12 \pm 544}{10} = \frac{12 \pm 23.32}{10}$ Positive root lo: $t = \frac{35.32}{10} \approx 3.53 s$ . Yeh step kyun? Do roots aate hain (ek negative "time before launch" jo unphysical hai); hum positive wala rakhte hain — real landing moment.

Step 3 — Horizontal distance. Horizontal speed constant hai: $u_{x} = 15 (0.6) = 9 m/s$ . $x = u_{x} t = 9 (3.53) \approx 31.8 m$ Yeh step kyun? Koi horizontal force nahi ⇒ poori flight mein uniform motion, cliff ho ya na ho.

Verify: flat-ground time hota $T = \frac{2 ( 15 ) ( 0.8 )}{10} = 2.4$ s; hamara cliff time $3.53$ s zyada hai — sahi, kyunki woh extra 20 m girta hai ✓. Discriminant $544 > 0$ to real landing exist karta hai ✓. $t = 3.53$ back plug karo: $12 (3.53) - 5 (3.53)^{2} = 42.4 - 62.3 \approx - 19.9 \approx - 20$ ✓.

Example 6 — Cell G: real-world word problem

Worked example Statement

Ek firefighter ki hose $u = 30 m/s$ par paani eject karti hai. $θ = 4 5^{\circ}$ par aim kiya gaya, kya jet $22 m$ door ek wall clear karega, aur wall tak pahunchte waqt paani kitna upar hai? ( $g = 10$ , $sin 4 5^{\circ} = cos 4 5^{\circ} \approx 0.7071$ .)

Forecast: $4 5^{\circ}$ maximum range deta hai. Guess karo kya $22$ m us reach ke andar comfortably hai.

Step 1 — English ko symbols mein translate karo. "Wall clear karna" matlab range $22$ m se zyada honi chahiye. $4 5^{\circ}$ par range: $sin 9 0^{\circ} = 1$ . $R = \frac{u ^{2} s i n 9 0 ^{\circ}}{g} = \frac{900}{10} = 90 m$ Yeh step kyun? $R$ batata hai paani ground level par kahan land karta hai ( $90$ m ≫ $22$ m), to woh wall ki horizontal position zaroor cross karta hai.

Step 2 — Wall tak pahunchne ka time nikalo. $u_{x} = 30 (0.7071) = 21.21 m/s$ . $t = \frac{x}{u _{x}} = \frac{22}{21.21} \approx 1.037 s$ Yeh step kyun? Hume $x = 22$ par height chahiye, to pehle woh time nikalo jab paani us horizontal spot par ho.

Step 3 — Us time par height. $u_{y} = 21.21 m/s$ . $y = u_{y} t - \frac{1}{2} g t^{2} = 21.21 (1.037) - 5 (1.037)^{2} \approx 21.99 - 5.38 \approx 16.6 m$ Yeh step kyun? Vertical position free-fall equation use karta hai; woh height batati hai ki jet wall ke top se upar hai ya nahi.

Verify: paani wall par $\approx 16.6$ m upar hai — us se neeche ki koi bhi wall clear ho jaayegi. Poore arc ki max height hai $H = \frac{900 ( 0.5 )}{20} = 22.5$ m, aur $16.6 < 22.5$ ✓ ( $x = 22$ par abhi peak nahi aaya, sensible kyunki peak $x = R /2 = 45$ m par hai). Units metres ✓.

Example 7 — Cell H: limiting behaviour (answers scale kaise karte hain)

Worked example Statement

$θ = 4 5^{\circ}$ fix rakho. (a) Agar launch speed $u$ se double hokar $2 u$ ho jaaye, to $R$ aur $H$ ka kya hoga? (b) Moon par gravity lagbhag $g /6$ hai. $R$ kaise change hoga?

Forecast: guess karo — kya speed double karne se range double hoti hai, ya zyada?

Step 1 — Speed scaling. $R$ aur $H$ dono mein $u^{2}$ hai: $\frac{R ( 2 u )}{R ( u )} = \frac{( 2 u ) ^{2}}{u ^{2}} = 4, \frac{H ( 2 u )}{H ( u )} = 4$ Yeh step kyun? Formulas $u$ par nahi, $u^{2}$ par depend karte hain — to speed par koi bhi factor answer mein square ho jaata hai. Speed double ⇒ range aur height chaar guna.

Step 2 — Gravity scaling. $R \propto \frac{1}{g}$ : $\frac{R _{moon}}{R _{earth}} = \frac{1/ ( g /6 )}{1/ g} = 6$ Yeh step kyun? Kamzor gravity matlab hawa mein zyada waqt ( $T \propto 1/ g$ ) aur range ko woh $1/ g$ virasat mein milta hai — Moon par same throw chhe guna door jaata hai.

Verify: $u = 20$ , $θ = 4 5^{\circ}$ ke saath concrete check. Earth: $R = \frac{400 ( 1 )}{10} = 40$ m. Speed double karke 40: $R = \frac{1600}{10} = 160$ m $= 4 \times 40$ ✓. Moon ( $g = 10/6 \approx 1.667$ ): $R = \frac{400}{1.667} \approx 240$ m $= 6 \times 40$ ✓.

Example 8 — Cell I: exam twist (do facts, hidden angle)

Worked example Statement

Ek projectile ki maximum height $H = 15 m$ aur horizontal range $R = 40 m$ hai ( $g = 10$ ). Launch angle $θ$ nikalo. (Tumhe $u$ nahi bataya gaya.)

Forecast: do unknowns ( $u, θ$ ), do facts. Guess karo: kya $H$ aur $R$ ka ratio $u$ magically cancel kar dega?

Step 1 — Dono formulas likho. $H = \frac{u ^{2} s i n ^{2} θ}{2 g}, R = \frac{u ^{2} s i n 2 θ}{g} = \frac{2 u ^{2} s i n θ c o s θ}{g}$ Yeh step kyun? Dono ko side by side rakhne se shared $u^{2}$ dikhta hai — ek clue ki divide karne par woh erase ho jaayega.

Step 2 — $H / R$ ka ratio lo taaki $u^{2}$ aur $g$ cancel ho jaayein. $\frac{H}{R} = \frac{u ^{2} s i n ^{2} θ / ( 2 g )}{2 u ^{2} s i n θ c o s θ / g} = \frac{s i n ^{2} θ}{4 s i n θ c o s θ} = \frac{s i n θ}{4 c o s θ} = \frac{t a n θ}{4}$ Yeh step kyun? Unknown speed gayab ho jaati hai, sirf $θ$ mein ek equation bachti hai. Bacha hua $\frac{s i n θ}{c o s θ}$ hi $tan θ$ hai — natural tool kyunki yeh poore angle ko ek ratio mein pack karta hai (dekho Vectors — resolving into components).

Step 3 — $θ$ ke liye solve karo. $\frac{15}{40} = \frac{t a n θ}{4} \Rightarrow tan θ = \frac{4 \cdot 15}{40} = 1.5 \Rightarrow θ = arctan (1.5) \approx 56. 3^{\circ}$ Yeh step kyun? $arctan$ answer karta hai "kis angle ka yeh tangent hai?" — launch angle recover karne ke liye tan ko undo karta hai.

Verify: $H$ se $u$ recover karo: $u^{2} = \frac{2 g H}{s i n ^{2} θ} = \frac{2 ( 10 ) ( 15 )}{s i n ^{2} 56. 3 ^{\circ}} = \frac{300}{0.692} \approx 433.6$ , to $u \approx 20.8$ m/s. $R$ mein feed karo: $R = \frac{433.6 s i n ( 112. 6 ^{\circ} )}{10} = \frac{433.6 ( 0.923 )}{10} \approx 40.0$ m ✓ — diye gaye range se match karta hai.

Recall

Complementary-angle twins

θ

aur

9 0^{\circ} - θ

equal range dete hain kyunki

sin 2 θ = sin (18 0^{\circ} - 2 θ)

Range box cliff se kyun use nahi kar sakte?

Launch aur landing heights alag hain;

R = \frac{u ^{2} s i n 2 θ}{g}

assume karta hai ki woh equal hain.

H

aur

R

diye jaane par unknown

u

kaise cancel karte ho?

H / R = \frac{t a n θ}{4}

banao, phir

θ = arctan (4 H / R)

Launch speed double karne par range kitne guna ho jaati hai?

Chaar guna (range

\propto u^{2}

Connections

Range, max height, time of flight — all derived (parent)
Projectile Motion — components & independence
Equations of Motion (1D kinematics)
Vectors — resolving into components
Free Fall & Acceleration due to gravity
Trigonometric identities — double angle
Energy method — max height via conservation