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Practice question

MCQ

Question

A projectile is launched at 25m/s at 37∘. What is its vertical velocity at t\=1s? (Take g\=10m/s2,sin⁡37∘\=0.6)

Options

Choose one · Correct answer highlighted

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 5 m/s. Hence option C satisfies projectile formulas.

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