Practice question
Question
A projectile is launched at 26m/s at 53∘. What is its horizontal velocity component? (Take cos53∘\=0.6)
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 12 m/s. Hence option A satisfies projectile formulas.
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