Intuition The big pictureA projectile is just two independent 1D motions glued together :
Horizontal: constant velocity (no force, ignoring air). Boring, steady.Vertical: free fall under g g g
The only thing connecting them is time t t t every projectile formula. We never memorize — we derive.
WHY split into components? Because force (gravity) acts only vertically. Horizontal motion feels no force, so it must be uniform. Splitting lets us use simple 1D equations on each axis.
WHAT: Total time in the air, from launch until it lands back at the same height.
HOW: Use vertical displacement y = u y t − 1 2 g t 2 y = u_y t - \tfrac{1}{2}g t^2 y = u y t − 2 1 g t 2 y = 0 y=0 y = 0
0 = u sin θ t − 1 2 g t 2 = t ( u sin θ − 1 2 g t ) 0 = u\sin\theta \, t - \tfrac{1}{2} g t^2 = t\left(u\sin\theta - \tfrac{1}{2}g t\right) 0 = u sin θ t − 2 1 g t 2 = t ( u sin θ − 2 1 g t )
Why this step? Factoring out t t t t = 0 t=0 t = 0
Intuition Symmetry insightTime to reach the top = u sin θ g = \dfrac{u\sin\theta}{g} = g u sin θ = = = = 2 × = 2\times = 2 ×
WHAT: Highest vertical point, where v y = 0 v_y = 0 v y = 0
HOW: Use v y 2 = u y 2 − 2 g H v_y^2 = u_y^2 - 2g\,H v y 2 = u y 2 − 2 g H v y = 0 v_y = 0 v y = 0
0 = ( u sin θ ) 2 − 2 g H 0 = (u\sin\theta)^2 - 2gH 0 = ( u sin θ ) 2 − 2 g H
Why this step? v y 2 = u y 2 − 2 g y v_y^2 = u_y^2 - 2gy v y 2 = u y 2 − 2 g y time-free energy-like relation — perfect because at the peak we know velocity (0 0 0
Worked example Cross-check using timeAt t = u sin θ g t = \tfrac{u\sin\theta}{g} t = g u s i n θ y = u y t − 1 2 g t 2 y = u_y t - \tfrac12 g t^2 y = u y t − 2 1 g t 2 H = u sin θ ⋅ u sin θ g − 1 2 g u 2 sin 2 θ g 2 = u 2 sin 2 θ g − u 2 sin 2 θ 2 g = u 2 sin 2 θ 2 g ✓ H = u\sin\theta\cdot\frac{u\sin\theta}{g} - \frac12 g\frac{u^2\sin^2\theta}{g^2} = \frac{u^2\sin^2\theta}{g} - \frac{u^2\sin^2\theta}{2g} = \frac{u^2\sin^2\theta}{2g}\ \checkmark H = u sin θ ⋅ g u s i n θ − 2 1 g g 2 u 2 s i n 2 θ = g u 2 s i n 2 θ − 2 g u 2 s i n 2 θ = 2 g u 2 s i n 2 θ ✓ Why this step? Two independent routes giving the same answer proves the derivation.
WHAT: Horizontal distance covered during the full flight T T T
HOW: Horizontal motion is uniform, so R = u x ⋅ T R = u_x \cdot T R = u x ⋅ T
R = u cos θ ⋅ 2 u sin θ g = 2 u 2 sin θ cos θ g R = u\cos\theta \cdot \frac{2u\sin\theta}{g} = \frac{2u^2\sin\theta\cos\theta}{g} R = u cos θ ⋅ g 2 u s i n θ = g 2 u 2 s i n θ c o s θ
Use the identity 2 sin θ cos θ = sin 2 θ 2\sin\theta\cos\theta = \sin 2\theta 2 sin θ cos θ = sin 2 θ
Why this step? Range = (constant horizontal speed) × (total time aloft). Time aloft is set entirely by vertical motion — that's the coupling.
45 ∘ 45^\circ 4 5 ∘ R ∝ sin 2 θ R \propto \sin 2\theta R ∝ sin 2 θ 2 θ = 90 ∘ ⇒ θ = 45 ∘ 2\theta = 90^\circ \Rightarrow \theta = 45^\circ 2 θ = 9 0 ∘ ⇒ θ = 4 5 ∘ Maximum range = u 2 g = \dfrac{u^2}{g} = g u 2 θ \theta θ 90 ∘ − θ 90^\circ-\theta 9 0 ∘ − θ same range because sin 2 θ = sin ( 180 ∘ − 2 θ ) \sin 2\theta = \sin(180^\circ - 2\theta) sin 2 θ = sin ( 18 0 ∘ − 2 θ )
Recall The 3 boxed resultsT = 2 u sin θ g , H = u 2 sin 2 θ 2 g , R = u 2 sin 2 θ g T = \frac{2u\sin\theta}{g}, \quad H = \frac{u^2\sin^2\theta}{2g}, \quad R = \frac{u^2\sin 2\theta}{g} T = g 2 u s i n θ , H = 2 g u 2 s i n 2 θ , R = g u 2 s i n 2 θ H H H sin 2 \sin^2 sin 2 R R R sin 2 θ \sin 2\theta sin 2 θ T T T sin \sin sin All vanish at θ = 0 \theta=0 θ = 0 θ = 0 \theta = 0 θ = 0 T = 0 , H = 0 , R = 0 T=0,H=0,R=0 T = 0 , H = 0 , R = 0
Worked example Example 1 — full kitA ball is kicked at u = 20 m/s u = 20\ \text{m/s} u = 20 m/s θ = 30 ∘ \theta = 30^\circ θ = 3 0 ∘ g = 10 m/s 2 g = 10\ \text{m/s}^2 g = 10 m/s 2
Time of flight: T = 2 ( 20 ) ( 0.5 ) 10 = 2 s T = \dfrac{2(20)(0.5)}{10} = 2\ \text{s} T = 10 2 ( 20 ) ( 0.5 ) = 2 s Why? sin 30 ∘ = 0.5 \sin 30^\circ = 0.5 sin 3 0 ∘ = 0.5
Max height: H = ( 20 ) 2 ( 0.5 ) 2 2 ( 10 ) = 400 ⋅ 0.25 20 = 5 m H = \dfrac{(20)^2(0.5)^2}{2(10)} = \dfrac{400\cdot 0.25}{20} = 5\ \text{m} H = 2 ( 10 ) ( 20 ) 2 ( 0.5 ) 2 = 20 400 ⋅ 0.25 = 5 m Why? Square the sine and the speed.
Range: R = ( 20 ) 2 sin 60 ∘ 10 = 400 ⋅ 0.866 10 ≈ 34.6 m R = \dfrac{(20)^2\sin 60^\circ}{10} = \dfrac{400\cdot 0.866}{10} \approx 34.6\ \text{m} R = 10 ( 20 ) 2 sin 6 0 ∘ = 10 400 ⋅ 0.866 ≈ 34.6 m Why? Use sin 2 θ = sin 60 ∘ \sin 2\theta = \sin 60^\circ sin 2 θ = sin 6 0 ∘ sin 30 ∘ \sin 30^\circ sin 3 0 ∘
Worked example Example 2 — Forecast-then-VerifySame speed, but θ = 60 ∘ \theta = 60^\circ θ = 6 0 ∘ Forecast: Range should be identical to Example 1 (complementary angles).
Verify: R = 400 sin 120 ∘ 10 = 400 ( 0.866 ) 10 ≈ 34.6 m R = \dfrac{400\sin 120^\circ}{10} = \dfrac{400(0.866)}{10} \approx 34.6\ \text{m} R = 10 400 sin 12 0 ∘ = 10 400 ( 0.866 ) ≈ 34.6 m H = 400 ( 0.75 ) 20 = 15 m H = \dfrac{400(0.75)}{20} = 15\ \text{m} H = 20 400 ( 0.75 ) = 15 m T = 2 ( 20 ) ( 0.866 ) 10 ≈ 3.46 s T = \dfrac{2(20)(0.866)}{10}\approx 3.46\ \text{s} T = 10 2 ( 20 ) ( 0.866 ) ≈ 3.46 s
sin θ \sin\theta sin θ sin 2 θ \sin 2\theta sin 2 θ Why it feels right: Every other formula uses sin θ \sin\theta sin θ The fix: Range comes from u cos θ × 2 u sin θ g u\cos\theta \times \frac{2u\sin\theta}{g} u cos θ × g 2 u s i n θ sin \sin sin cos \cos cos collapses into sin 2 θ \sin 2\theta sin 2 θ 2 sin θ cos θ 2\sin\theta\cos\theta 2 sin θ cos θ
Common mistake Forgetting the factor of 2 in time of flightWhy it feels right: t up = u sin θ g t_\text{up} = \frac{u\sin\theta}{g} t up = g u s i n θ The fix: That's only the rise time. Total flight = up and down = 2 t up 2t_\text{up} 2 t up t up t_\text{up} t up
Common mistake Treating horizontal velocity as changingWhy it feels right: "Everything slows down eventually."
The fix: No horizontal force ⇒ u x u_x u x u y u_y u y
Common mistake Range formula used when launch and landing heights differWhy it feels right: It's the formula you memorized.
The fix: R = u 2 sin 2 θ g R = \frac{u^2\sin2\theta}{g} R = g u 2 s i n 2 θ same height . Off a cliff, re-derive T T T y = 0 y=0 y = 0
Recall Explain to a 12-year-old (click to reveal)Imagine throwing a ball. Two things happen at the same time but separately : it travels forwards at a steady pace (nothing pushes it sideways), and it goes up then down because gravity pulls it back like an invisible rubber band. The time it spends in the air depends only on the up-down part. While it's floating there, the forward part keeps moving steadily. So:
Time in air = how long the up-down trip takes.How high = how strong the upward throw was.How far = forward speed × time in air.
Throwing at a slant of exactly 45° sends it farthest, because you balance "going up long enough" with "going forward fast."
Mnemonic Remember the powers ofu u u g g g "Time is fair, Height is steep, Range is loose."
T ime: u 1 , g 1 u^1, g^1 u 1 , g 1 H eight: u 2 , g 1 u^2, g^1 u 2 , g 1 2 (steep climb costs).R ange: u 2 , g 1 u^2, g^1 u 2 , g 1 sin 2 θ \sin 2\theta sin 2 θ
What two motions make up projectile motion, and what links them? Horizontal (uniform velocity) and vertical (free fall under
g g g ); linked only by the shared time
t t t .
Derive time of flight from y = u sin θ t − 1 2 g t 2 = 0 y = u\sin\theta\, t - \frac12 g t^2 = 0 y = u sin θ t − 2 1 g t 2 = 0 Factor
t ( u sin θ − 1 2 g t ) = 0 t(u\sin\theta - \frac12 g t)=0 t ( u sin θ − 2 1 g t ) = 0 ; nonzero root
T = 2 u sin θ g T = \frac{2u\sin\theta}{g} T = g 2 u s i n θ .
Formula for time of flight? T = 2 u sin θ g T = \frac{2u\sin\theta}{g} T = g 2 u s i n θ Formula for maximum height? H = u 2 sin 2 θ 2 g H = \frac{u^2\sin^2\theta}{2g} H = 2 g u 2 s i n 2 θ Formula for horizontal range? R = u 2 sin 2 θ g R = \frac{u^2\sin 2\theta}{g} R = g u 2 s i n 2 θ Why does max range occur at 45°? R ∝ sin 2 θ R\propto\sin 2\theta R ∝ sin 2 θ , maximized when
2 θ = 90 ° 2\theta=90° 2 θ = 90° , i.e.
θ = 45 ° \theta=45° θ = 45° ; max range
= u 2 / g =u^2/g = u 2 / g .
Why do complementary angles give the same range? sin 2 θ = sin ( 180 ° − 2 θ ) \sin 2\theta = \sin(180°-2\theta) sin 2 θ = sin ( 180° − 2 θ ) , so
θ \theta θ and
90 ° − θ 90°-\theta 90° − θ give equal
R R R .
At max height, what is the vertical velocity? Zero (it momentarily stops rising); horizontal velocity is still
u cos θ u\cos\theta u cos θ .
Which trig identity converts 2 sin θ cos θ 2\sin\theta\cos\theta 2 sin θ cos θ 2 sin θ cos θ = sin 2 θ 2\sin\theta\cos\theta = \sin 2\theta 2 sin θ cos θ = sin 2 θ .
For u = 20 , θ = 30 ° , g = 10 u=20,\theta=30°,g=10 u = 20 , θ = 30° , g = 10 T = 2 T=2 T = 2 s,
H = 5 H=5 H = 5 m,
R ≈ 34.6 R\approx 34.6 R ≈ 34.6 m.
Horizontal ux = u cos theta
Vertical uy = u sin theta
Time of flight T = 2u sin theta / g
Max height H = u squared sin sq theta / 2g
Range R = u squared sin 2theta / g
Identity 2 sin cos = sin 2theta
Intuition Hinglish mein samjhoDekho, projectile motion ko samajhne ka sabse asaan tareeka yeh hai: ball ke do alag-alag motion chal rahe hain ek saath. Ek horizontal motion jisme koi force nahi (gravity sirf neeche khichti hai), isliye horizontal speed u cos θ u\cos\theta u cos θ vertical motion jisme gravity kaam karti hai, toh ball pehle upar jaati hai, rukti hai, phir neeche aati hai. In dono ko jodne wali cheez sirf time t t t
Time of flight nikaalne ke liye vertical displacement zero rakho (kyunki ball wapas zameen pe aati hai): T = 2 u sin θ g T = \frac{2u\sin\theta}{g} T = g 2 u s i n θ 2 — kyunki upar jaane aur neeche aane dono ka time hai, sirf upar ka nahi. Max height wahan hai jahan vertical velocity zero ho jaati hai: H = u 2 sin 2 θ 2 g H = \frac{u^2\sin^2\theta}{2g} H = 2 g u 2 s i n 2 θ Range toh simply horizontal speed × total time = u 2 sin 2 θ g \frac{u^2\sin2\theta}{g} g u 2 s i n 2 θ sin θ \sin\theta sin θ cos θ \cos\theta cos θ sin 2 θ \sin 2\theta sin 2 θ
Ek important baat: maximum range 45° pe milti hai, kyunki sin 2 θ \sin 2\theta sin 2 θ 30 ° 30° 30° 60 ° 60° 60° sin 2 θ \sin 2\theta sin 2 θ sin θ \sin\theta sin θ 2 u 2 sin θ cos θ g \frac{2u^2\sin\theta\cos\theta}{g} g 2 u 2 s i n θ c o s θ
