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JEE Main 2021 (27 Aug, Shift 2) · Time of Flight, Maximum Height & Range

Time of Flight, Maximum Height & Range question from JEE Main 2021 (27 Aug, Shift 2)

Q25JEE Main 2021 · 27 AUG · SHIFT 2Easy

Time of Flight, Maximum Height & Range

A player kicks a football with an initial speed of 25 ms−1{}^{-1} at an angle of 45∘^\circ from the ground. What are the maximum height and the time taken by the football to reach at the highest point during motion? (Take g = 10 ms−2{}^{-2})

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Answer: (c)

  1. H=u2sin⁡2θ2g=625×1220=15.625 mH = \frac{u^2\sin^2\theta}{2g} = \frac{625 \times \frac{1}{2}}{20} = 15.625\text{ m}
  2. Time to reach the top: t=usin⁡θg=25×0.70710≈1.77 st = \frac{u\sin\theta}{g} = \frac{25 \times 0.707}{10} \approx 1.77\text{ s}
  3. So hmax=15.625 mh_{max} = 15.625\text{ m}, t=1.77 st = 1.77\text{ s}, answer (c).
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