Projectile Motion

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1. CSCA Physics December 2025 Past Paper
   A small ball is thrown horizontally from a building of height $h = 20 \mathrm {~m}$ with an initial speed of $v _ { 0 } = 20 \mathrm {~m} / \mathrm { s }$. Neglecting air resistance, the magnitude of the ball's velocity when it hits the ground is ( ) $\left( g = 10 \mathrm {~m} / \mathrm { s } ^ { 2 } \right)$
   • A．$40 \sqrt { 2 } \mathrm {~m} / \mathrm { s }$
   • B．$20 \sqrt { 2 } \mathrm {~m} / \mathrm { s }$
   • C．$40 \mathrm {~m} / \mathrm { s }$
   • D．$20 \mathrm {~m} / \mathrm { s }$
   Antwort & Lösungsweg anzeigen

   Antwort: B

   1 Horizontal projectile: time to fall $t = \sqrt { 2 h / g } = \sqrt { 2 \times 20 / 10 } \mathrm {~s} = 2 \mathrm {~s}$. 2 Horizontal speed stays $20 \mathrm {~m} / \mathrm { s }$; vertical speed at impact $v _ { y } = g t = 20 \mathrm {~m} / \mathrm { s }$. 3 Resultant speed $v = \sqrt { 20 ^ { 2 } + 20 ^ { 2 } } = 20 \sqrt { 2 } \mathrm {~m} / \mathrm { s }$, option B.
   In der vollständigen CSCA Physics December 2025 Past Paper ansehen
2. CSCA Physics March 2026 Past Paper
   If a projectile's initial velocity is tripled, then
   • A．Flight time triples
   • B．Horizontal displacement triples
   • C．Landing speed unchanged
   • D．Vertical displacement triples
   Antwort & Lösungsweg anzeigen

   Antwort: B

   Flight time $t = \sqrt { 2 h / g }$ : depends on height only (A wrong). Vertical displacement depends on $t$ and $g$ only (D wrong). $\mathrm { x } = \mathrm { vt }$ : tripling v triples $\mathrm { x } ( \mathrm { B } )$. Vertical component unchanged but horizontal increases → landing speed increases (C wrong). Answer: B.
   Die komplette CSCA Physics March 2026 Past Paper bearbeiten – mit Zeit & Bewertung