Skip to content
Login Register

Practice question

MCQ

Question

A projectile is launched at 16m/s at 37∘. What is its horizontal velocity component? (Take cos⁡37∘\=0.8)

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 10 m/s. Hence option A satisfies projectile formulas.

Related blogs

Guides and insights connected to this topic