Practice question
Question
A projectile is launched at 14m/s at 53∘. What is its time of flight? (Take g\=9.8m/s2,sin53∘\=0.8)
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 2 v. Hence option A satisfies projectile formulas.
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