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A player kicks a football with an initial speed of $25 \mathrm{~ms}^{-1}$ 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 ? $\left ( Take\, \: g=10ms^{-2} \right )$

Option: 1 $\begin{aligned} &h_{\max }=3.54 \mathrm{~m} \mathrm{~T}=0.125 \mathrm{~s} \\ \end{aligned}$
Option: 2 $h_{\max }=15.625 \mathrm{~m} \mathrm{} \mathrm{T}=1.77 \mathrm{~s} \\$
Option: 3 $h_{\max }=15.625 \mathrm{~m} \mathrm{~T}=3.54 \mathrm{~s} \\$
Option: 4 $h_{\max }=10 \mathrm{~m} \mathrm{} \mathrm{T}=2.5 \mathrm{~s}$

Answers (1)

best_answer

$u = 25m/s\\$

$\theta= 45^{\circ}\\$

The maximum height reached is

$H= \frac{u \sin^{2}\theta}{2g}= \frac{\left ( 25 \right )^{2}\times\left ( \sin45 \right )^{2}}{2\times10}\\$

$= \frac{625\times\frac{1}{2}}{20}\\$

$= \frac{625}{40}\\$

$H= 15.625\\$

Time taken to reach maximum height is $\frac{1}{2}\left ( T_{flight} \right )\\$

$t= \frac{1}{2}\times\frac{2u\sin\theta}{g}\\$

$= \frac{25\times\frac{1}{\sqrt{2}}}{10}= 2.5\times0.7\\$

$t\cong 1.77s$

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vishal kumar

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