Question bank — Projectile motion — horizontal - vertical independence, full derivation

False — sirf vertical velocity $v_y$ wahan zero hoti hai; acceleration poori flight mein $g$ downward rehti hai, kyunki gravity kabhi band nahi hoti.

False — $a_x=0$, isliye $v_x=u\cos \theta$ constant rehta hai. Total speed isliye girta hai kyunki $v_y$ kam hoti hai, na ki sideways motion khatam hone ki wajah se.

False — $a_y=-g$ mein koi mass nahi hai, isliye mass cancel ho jaata hai; air drag ignore karne par, range sirf $u$, $\theta$, $g$ par depend karti hai, weight par nahi.

False — $R=\frac{u^2\sin 2\theta }{g}$ mein $u^2$ ke saath scale hota hai, isliye $u$ double karne se range chaar guna ho jaati hai.

Aadha sach hai — ranges equal hain (complementary angles), lekin $T\propto \sin \theta$, isliye $60^{\circ }$ wala shot zyada der tak hawaai mein rehta hai.

False — ye sirf seedha upar phenkne ke liye sach hai. General projectile mein $v_x=u\cos \theta \ne 0$, isliye wo top par bhi sideways cruise karta rehta hai.

True — vertical motion top ke baare mein symmetric hai: $t_{\text{up}}=t_{\text{down}}=\frac{u\sin \theta }{g}=T/2$, kyunki gravity constant hai.

False — dono ki vertical motion identical hai ($u_y=0$, same $g$, same height), isliye dono saath mein land karte hain; horizontal travel ka landing time se koi lena-dena nahi.

False — range sirf $45^{\circ }$ tak badhti hai, phir wapas girti hai; $\sin 2\theta$ peak karta hai jab $2\theta =90^{\circ }$ ho.

True — $v_x$ unchanged rehta hai aur $|v_y|$ landing par symmetry se $u\sin \theta$ ke barabar hota hai, isliye total speed $=u$ (air drag ignore karne par).

Galat — ye formula assume karta hai ki landing launch height par ho. Cliff ke liye, pehle poori $y=u_{y}t-\frac{1}{2}g{t}^2$ solve karo $t$ ke liye, phir $x=u_{x}t$.

Do galtiyan hain: $v_x$ $u\cos \theta$ hai aur kabhi zero nahi hota, aur gravity $v_y$ par act karti hai, $v_x$ par nahi. Top woh jagah hai jahan $v_y=u\sin \theta -gt=0$.

Galat — sirf $v_y$ kam hota hai; $v_x$ constant hai, isliye ye equal horizontal distances equal times mein cover karta hai poori flight mein.

Sign error — upar positive hai aur gravity neeche kheenchti hai, isliye ye $-\frac{1}{2}g{t}^2$ hona chahiye. Plus sign se ball hamesha upar jaati rahegi.

Galat component — flight time vertical motion se aati hai, isliye ye $\sin \theta$ use karta hai: $T=\frac{2u\sin \theta }{g}$.

Formula zero deta hai kyunki ye ground level par instantly land ho jaata hai — ye level ground ke liye sahi hai. Ek uthe hue platform se ye zaroor travel karta hai; formula wahan apply nahi hota.

Kyunki $t$ eliminate karne par milta hai $y=x\tan \theta -\frac{g{x}^2}{2u^2{\cos }^2\theta }$, jo $x$ mein quadratic hai ($y=ax-b{x}^2$) — parabola ki defining shape.

Newton's law per component hold karta hai aur gravity sirf $a_y$ mein enter hoti hai; dono equations mein sirf common clock $t$ share hoti hai, isliye ye decoupled hain — dekho Newton's Second Law.

$R\propto \sin 2\theta$, aur $\sin 2\theta$ sabse bada ($=1$) hota hai jab $2\theta =90^{\circ }$, yaani $\theta =45^{\circ }$ — horizontal reach aur airtime ke beech ka sweet spot.

Velocity arrow ki shadow x-axis par (adjacent side) $u\cos \theta$ hai aur y-axis par (opposite side) $u\sin \theta$ hai — ye Vectors — Resolving into Components se pure right-triangle geometry hai.

Kyunki $a=\frac{dv}{dt}$ aur $v=\frac{dx}{dt}$ — integrate karna derivative ko undo karta hai, known acceleration ko velocity mein aur phir position mein convert karta hai (dekho Calculus — Integration).

Vertical motion symmetric hai, isliye $|v_y|$ wapas $u\sin \theta$ par aa jaata hai jabki $v_x$ unchanged rehta hai; velocity triangle mirror ho jaata hai, same angle magnitude deta hai.

$u\cos 90^{\circ }=0$, isliye koi horizontal motion nahi — "parabola" ek vertical line mein degenerate ho jaati hai upar aur neeche; range zero hai.

Level ground ke liye zero — object pehle se $y=0$ par hai aur kabhi upar nahi uthta, isliye $t=0$ par "land" kar leta hai; travel dekhne ke liye nonzero launch height chahiye.

Sab kam ho jaate hain: $T\propto 1/g$ (aadha), $H\propto 1/g$ (aadha), $R\propto 1/g$ (aadha) — zyada strong gravity use jaldi neeche kheench leti hai. Compare karo Free Fall and g se.

Range badhti hai kyunki $R\propto 1/g$; weak gravity matlab zyada airtime aur zyada door flight — same trajectory shape, bas stretched.

Purely horizontal, magnitude $u\cos \theta$ ke saath; vertical part abhi-abhi upar se neeche jaate waqt zero cross kiya hai, lekin arrow poori tarah kabhi gayab nahi hota.

Nahi — wind ek horizontal force add karti hai, isliye $a_x\ne 0$; horizontal equation ab constant-velocity nahi rehti, haalaanki vertical free fall rehti hai. Ye Relative Velocity se connect hota hai.