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Answer: 4. \(20\sqrt{2}\) m/s, 10 m/s\(^2\)

Key idea: once the ball is dropped, its horizontal velocity stays equal to the car's velocity at that instant, and only gravity acts (acceleration 10 m/s\(^2\) downward).

  • At t = 4 s car speed \(=5\times4=20\) m/s, so \(v_x=20\) m/s.
  • In the next 2 s: \(v_y=gt=10\times2=20\) m/s.

\(v=\sqrt{20^2+20^2}=20\sqrt2\) m/s, acceleration \(=g=10\) m/s\(^2\). Option 4.

Doubtly AI · 30 Oct 2026 AI-generated · not yet verified Report
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