Horizontal Velocity Component: 18 m/s at 45°

Practice question

Question

A projectile is launched at 18m/s at 45∘. What is its horizontal velocity component?

Options

Choose one · Correct answer highlighted

10 m/s Correct 18 m/s 8 m/s 12 m/s

Explanation

Projectile motion splits into horizontal uniform and vertical accelerated motion per NCERT Chapter 4. Range R=u²sin2θ/g and max height H=u²sin²θ/2g. Using given u, θ, g, calculation yields 10 m/s. Hence option A satisfies projectile formulas.

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