Projectile Motion

How the real exam tests it

Questions from real CSCA papers that test this concept. Work each one out before you open the answer.

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．$20 \mathrm {~m} / \mathrm { s }$
   Show answer & solution

   Answer: 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.
   See it in the full CSCA Physics December 2025 Past Paper
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
   Show answer & solution

   Answer: 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.
   Take the full CSCA Physics March 2026 Past Paper — timed & graded