Kinematics

JEE Main level, single correct, medium

251 of 308

Question 251 of 308

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

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Answer: (A) hmax⁡=15.625m,T=1.77s{h}_{\max }=15.625m,T=1.77s

θ=45∘\theta =45^{\circ}
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}
=usin⁡θg=25(12)10=25102=522s=\dfrac{u\sin \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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