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During a football match, the ball kicked at 45° angle of elevation went just over the goal post, height 2.4m. Assuming the goal post height is the greatest, calculate: The speed at which the ball was projected, The time taken to reach the greatest height

the horizontal distance between the point of kick and foot of the goal post bar (neglect the thickness of the bar)​

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Answer:

see below

Step-by-step explanation:

We are here given that ,

  • maximum height of projectile= 2.4m (h)
  • angle at which it is kicked = 45°
  • speed of projection= ? (u)

As we know that,


\implies h =(u^2\sin^2\theta)/(2g) \\

substitute the respective values,


\implies 2.4 m =(u^2\sin^245^\circ)/(2* 10) \\


\implies 2.4m =(u^2* \bigg((1)/(\sqrt2)\bigg)^2)/(20)\\


\implies u^2 = 40 * 2.4 = 96 \\


\implies u^2 =√(96)\\


\implies \underline{\underline{ u \approx 9.79 \ m/s }}\\

secondly we know that,


\implies t = (u\sin\theta)/(g)\\


\implies t =(9.79* \sin45^\circ)/(10) \\


\implies t =(9.79* (1)/(\sqrt2))/(10) \\


\implies \underline{\underline{ t \approx 0.69 \ s }}\\

and we are done!

User Guy Smorodinsky
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