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JEE Main 202127 AugustEvening Shift
Exam
JEE Main
Year
2021
Date
27 August
Shift
Evening Shift
Subject
Physics
Chapter
Motion in Two Dimensions

JEE Main 2021 · 27 August · Evening Shift · Physics · Motion in Two Dimensions

A player kicks a football with an initial speed of 25 ms ^ - 1 at...

A player kicks a football with an initial speed of 25ms−125{ms}^{- 1} at an angle of 45°45 ° 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=10ms−2g = 10{ms}^{- 2})

Options

  1. A{h}_\max = 15 . 625m, T = 1 . 77s
  2. B{h}_\max = 3 . 54m, T = 0 . 125s
  3. C{h}_\max = 10m, T = 2 . 5s
  4. D{h}_\max = 15 . 625m, T = 3 . 54s

Correct answer

A. {h}_\max = 15 . 625m, T = 1 . 77s

Step-by-step solution

θ=45°\theta = 45 °

H=u2sin⁡2θ2g=(25)2×(12)2×10=1258mH = \dfrac{{u}^{2}\sin^{2}\theta}{2g} = \dfrac{( 25{)}^{2} \times \left(\dfrac{1}{2}\right)}{2 \times 10} = \dfrac{125}{8}\text{m}

and time t=T2t = \dfrac{T}{2}

=usinθg=25(12)10=25102=522s= \dfrac{usin\theta}{g} = \dfrac{25\left(\dfrac{1}{\sqrt{2}}\right)}{10} = \dfrac{25}{10\sqrt{2}} = \dfrac{5}{2\sqrt{2}}\text{s}

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