In a circus,a performer throws an apple towar | vedclass

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(B) Let the point of projection be $A$ and the initial position of the hoop be $C$. The horizontal distance is $AB = 25 \, m$ and the vertical height

Vedclass · Accepted Answer

$22$

Vedclass · Answer

(B) Let the point of projection be $A$ and the initial position of the hoop be $C$. The horizontal distance is $AB = 25 \, m$ and the vertical height is $BC = 45 \, m$.The distance $AC = \sqrt{AB^2 + BC^2} = \sqrt{25^2 + 45^2} = \sqrt{625 + 2025} = \sqrt{2650} \, m$.The apple is thrown towards $C$ with speed $v = 24 \, m/s$. The time taken for the apple to reach the hoop's initial position $C$ is $t = \frac{AC}{v} = \frac{\sqrt{2650}}{24} \, s$.During this time $t$,the hoop falls vertically downwards by a distance $h = \frac{1}{2} g t^2$.Taking $g = 10 \, m/s^2$,we have $h = \frac{1}{2} 	imes 10 	imes \left(\frac{\sqrt{2650}}{24}ight)^2 = 5 	imes \frac{2650}{576} = \frac{13250}{576} \approx 23 \, m$.The height of the hoop above the ground when the apple passes through it is $H = 45 - h = 45 - 23 = 22 \, m$.

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