Range, max height, time of flight — all derived: Question bank

1.1.20 · D5 · Physics › Measurement, Vectors & Kinematics › Range, max height, time of flight — all derived

Full derivations ke liye parent dekho: Range, max height, time of flight — all derived.

True or false — justify

Har line: ek claim, phir ::: phir verdict reasoning ke saath.

Launch speed

u

ko double karne se range bhi double ho jaati hai.

False. Range mein

u^{2}

hai, isliye

u

ko double karne se

R

chaar guna ho jaata hai (

\frac{u ^{2} s i n 2 θ}{g}

se). Speed ka effect linear intuition se kahin zyada hota hai.

u

ko double karne se time of flight bhi double ho jaata hai.

True.

T = \frac{2 u s i n θ}{g}

u

mein linear hai, isliye same angle par speed double karne se airtime bhi double ho jaata hai.

Highest point par projectile ki speed zero hoti hai.

False. Top par sirf vertical velocity zero hoti hai. Horizontal velocity

u cos θ

kabhi nahi badlti, isliye wahan total speed

u cos θ

hoti hai (zero nahi, jab tak seedha upar launch na ho).

Same speed se

3 0^{\circ}

aur

6 0^{\circ}

par fire karne se same range milti hai.

True. Ye complementary hain (

3 0^{\circ} + 6 0^{\circ} = 9 0^{\circ}

), aur

sin 2 θ = sin (18 0^{\circ} - 2 θ)

, isliye

sin 6 0^{\circ} = sin 12 0^{\circ}

. Range same hogi, lekin

6 0^{\circ}

wala shot zyada upar aur zyada der tak jaayega.

Wo do complementary shots hawa mein bhi utna hi waqt bitaate hain.

False. Time

sin θ

par depend karta hai,

sin 2 θ

par nahi. Steep wale

6 0^{\circ}

shot ka

sin θ

zyada bada hai, isliye woh zyada der tak upar rehta hai, chahe same jagah utarta ho.

4 5^{\circ}

par launch kiya gaya projectile hamesha apni maximum possible height tak pahunchta hai.

False. Maximum range

4 5^{\circ}

par hoti hai; maximum height

9 0^{\circ}

par hoti hai (seedha upar), jahan

sin^{2} θ = 1

4 5^{\circ}

par height sirf

\frac{u ^{2}}{4 g}

hoti hai, seedha-upar wali maximum

\frac{u ^{2}}{2 g}

ki aadhi.

Agar

g

aadha kar do (jaise Moon-jaisi duniya par), toh range double ho jaati hai.

True.

R \propto \frac{1}{g}

, isliye

g

aadha karne se

R

double ho jaati hai. Kamzor gravity airtime bhi double karti hai, aur isliye ball aage zyada jaati hai.

Projectile ke uthne ke saath saath horizontal velocity kam hoti jaati hai.

False. Koi horizontal force nahi lagti (air ignored), isliye

u cos θ

poori flight mein constant rehti hai. Sirf vertical velocity badlti hai.

Upar jaane ka time neeche aane ke time ke barabar hota hai.

True (flat ground, no air). Gravity constant hai, isliye vertical motion time-mirror hai: upar jaate waqt decelerate karta hai aur neeche aate waqt same amount se re-accelerate karta hai, symmetrically.

Maximum height, vertical launch speed ke square ke proportional hoti hai.

True.

H = \frac{( u s i n θ ) ^{2}}{2 g} = \frac{u _{y}^{2}}{2 g}

. Ye sirf vertical component

u_{y} = u sin θ

ki parwah karta hai; horizontal part se koi fark nahi padta ki ball kitni upar jaayegi.

Spot the error

Har line mein ek galat step batayi gayi hai; reveal mein error ka naam aur fix bataya gaya hai.

"Range

= u cos θ \times \frac{u s i n θ}{g}

Jo time use kiya gaya woh sirf rise time

\frac{u s i n θ}{g}

hai, poora flight nahi. Total time

T = \frac{2 u s i n θ}{g}

hota hai — factor of

2

missing hai. Sahi range isko double karke

\frac{u ^{2} s i n 2 θ}{g}

deta hai.

"Range ke liye, angle seedha plug karo:

R = \frac{u ^{2} s i n θ}{g}

Product

2 sin θ cos θ

sin 2 θ

mein collapse hota hai,

sin θ

mein nahi.

sin θ

use karne se

cos θ

factor bilkul hi chhoot jaata hai — ye galti se predict karega ki maximum range

9 0^{\circ}

par hogi.

"Max height:

H = \frac{u s i n ^{2} θ}{2 g}

Speed ko bhi square karna zaroori hai:

H = \frac{u ^{2} s i n ^{2} θ}{2 g}

. Ye

v_{y}^{2} = u_{y}^{2} - 2 g H

se aata hai jahan

u_{y} = u sin θ

hai, isliye sine aur

u

dono square hote hain.

u = 20

m/s horizontal speed se cliff se roll kiya gaya ball ka range

\frac{u ^{2} s i n ( 2 \cdot 0 ^{\circ} )}{g} = 0

hai."

Flat-ground formula assume karta hai ki launch aur landing heights equal hain. Cliff se ye alag hoti hain, isliye tumhe

T

ko vertical drop

h = \frac{1}{2} g T^{2}

se re-derive karna hoga, phir

R = u_{x} T

. Ball clearly aage zaroor jaati hai.

θ = 9 0^{\circ}

par range maximum hai kyunki ball sabse zyada der tak udti hai."

Lamba airtime sirf aadhi baat hai:

9 0^{\circ}

par horizontal speed

u cos 9 0^{\circ} = 0

hai, isliye

R = 0

. Range ke liye dono airtime aur forward speed chahiye — balance point

4 5^{\circ}

hai.

"Kyunki

T

H

, aur

R

sab mein

sin

hai, isliye sab sirf

θ = 9 0^{\circ}

par zero hote hain."

Ye sab

θ = 0^{\circ}

par zero hote hain (flat ground par throw → koi airtime nahi).

R

9 0^{\circ}

par bhi zero hota hai (koi forward speed nahi), jabki

T

aur

H

wahan sabse bade hote hain. Alag quantities ke alag zeros hain.

"Energy se pata chalta hai ki max height launch angle par depend karti hai, isliye

H \propto sin θ

Energy argument deta hai

m g H = \frac{1}{2} m u_{y}^{2}

, isliye

H = \frac{u _{y}^{2}}{2 g} \propto sin^{2} θ

, na ki

sin θ

. Height mein vertical speed ka square convert hota hai, na ki speed khud.

Why questions

Time of flight horizontal launch speed par kyun independent hota hai?

Kyunki flight tab khatam hoti hai jab vertical motion ground level par wapas aati hai, aur vertical motion sirf gravity feel karta hai — horizontal component us clock ko na fast karta hai, na slow.

Range formula mein

sin θ cos θ

directly ki jagah

sin 2 θ

kyun use hota hai?

Dono equal hain: identity

2 sin θ cos θ = sin 2 θ

bas "kaafi der upar rehna" (

sin θ

) aur "aage tezi se jaana" (

cos θ

) ke product ko ek clean expression mein repackage karta hai jiska

4 5^{\circ}

par peak obvious hai.

4 5^{\circ}

range maximize kyun karta hai par height nahi?

Range airtime aur forward speed ke beech trade-off karta hai, balanced hota hai jab

2 θ = 9 0^{\circ}

yaani

4 5^{\circ}

. Height forward speed ko bilkul ignore karti hai aur bas maximum upward speed chahti hai, jo

9 0^{\circ}

par sabse badi hoti hai.

Hum horizontal aur vertical motion ko alag-alag kyun treat kar sakte hain?

Ek maatra force (gravity) purely vertically point karti hai, isliye sirf vertical velocity badlti hai. Horizontal aur vertical equations mein koi shared force term nahi hota — ye sirf shared time

t

ke through couple hote hain.

Seedha upar throw karne se range zero kyun hoti hai jabki woh sabse zyada der hawa mein rehti hai?

θ = 9 0^{\circ}

par horizontal component

u cos 9 0^{\circ} = 0

hota hai: kuch bhi ise sideways nahi le jaata, isliye yeh exactly wahi utarti hai jahan se gayi thi. Forward speed ke bina airtime se koi horizontal distance nahi milta.

Peak exactly aadhe total flight time par kyun pahuncha jaata hai?

Vertical velocity

u sin θ

0

tak linearly girta hai (uthte waqt) phir

- u sin θ

tak (girte waqt) constant

g

ke under. Zero velocity — peak — us symmetric linear graph ke midpoint par hota hai, isliye

t_{peak} = T /2

Kamzor gravity height aur range dono kyun badhati hai?

Chhota

g

matlab dheema vertical deceleration, isliye ball zyada upar uthti hai aur zyada der tak hawa mein rehti hai; extra airtime, unchanged horizontal speed ke saath, use aage bhi zyada le jaata hai.

H

aur

R

dono mein

\frac{1}{g}

hota hai.

Edge cases

θ = 0^{\circ}

par

T

H

R

ka kya hota hai?

Teeno zero hain:

sin 0 = 0

. Zameen se bilkul horizontal launch hone par, ball already landing height par hai, isliye koi airtime ya rise nahi hota — woh kabhi airborne nahi hoti.

θ = 9 0^{\circ}

(seedha upar) par kya hota hai?

T = \frac{2 u}{g}

(sabse lamba airtime) aur

H = \frac{u ^{2}}{2 g}

(possible sabse zyada height), lekin

R = 0

kyunki

sin 18 0^{\circ} = 0

— koi horizontal component nahi jo ise kahi le jaaye.

Fixed speed ke liye maximum height dene wala launch angle kaun sa hai?

θ = 9 0^{\circ}

H = \frac{u ^{2} s i n ^{2} θ}{2 g}

tab sabse bada hota hai jab

sin^{2} θ = 1

. Seedha upar saari speed vertical motion mein laga deta hai, deta hai

H_{max} = \frac{u ^{2}}{2 g}

Kya flat-ground formulas tab bhi chalte hain jab ball raised platform par ute?

Nahi. Ye equal launch aur landing heights assume karte hain. Agar landing point zyada upar (ya neeche) hai, toh tumhe vertical equation

y = u sin θ t - \frac{1}{2} g t^{2}

ko actual final

y

ke liye re-solve karna hoga,

y = 0

ke liye nahi.

Agar

θ

0^{\circ}

se thoda sa zyada ho, jaise

1^{\circ}

, toh range kya hogi?

Bahut chhoti par nonzero:

R = \frac{u ^{2} s i n 2 ^{\circ}}{g}

. Ball barely zameen chhodi aur almost turant land kar gayi — bahut chhoti airtime wala nearly grazing shot.

Agar do launches same range dete hain, kya unki launch speed same honi chahiye?

Zaroori nahi — ye traps equal

u

assume karte hain, lekin ek faster, steeper shot ek slower, flat shot ki range match kar sakta hai. Same speed se equal range hone par angles complementary hone chahiye; alag speeds us link ko tod deti hain.

Landing ke waqt (flat ground par), speed launch speed se kaise compare hoti hai?

Magnitude mein barabar hoti hain. Time-symmetry se vertical speed

u sin θ

(ab neeche ki taraf) wapas aa jaati hai jabki horizontal speed

u cos θ

rehti hai, isliye landing speed

u_{x}^{2} + u_{y}^{2} = u

hoti hai.

Recall Ek-line self-test (reveal karne ke liye click karo)

Agar koi tumhe bole "range $4 5^{\circ}$ par maximize hoti hai, isliye height bhi," toh sabse fast rebuttal kya hai? ::: Height forward speed ko ignore karti hai aur $sin^{2} θ$ ke saath badhti hai, $9 0^{\circ}$ par peak karti hai; $4 5^{\circ}$ result specifically airtime aur forward speed ke beech range trade-off hai.

Connections

Range, max height, time of flight — all derived
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