In a sports event,a disc is thrown such that | vedclass

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(B) For maximum range,the angle of projection is $	heta = 45^{\circ}$.The maximum range is given by $R_{\max} = \frac{u^2}{g} = 80 \ m$.Given $g = 10

Vedclass · Accepted Answer

$60$

Vedclass · Answer

(B) For maximum range,the angle of projection is $	heta = 45^{\circ}$.The maximum range is given by $R_{\max} = \frac{u^2}{g} = 80 \ m$.Given $g = 10 \ m/s^2$,we have $u^2 = 80 	imes 10 = 800$,so $u = \sqrt{800} \ m/s$.The horizontal distance travelled in time $t$ is $x = (u \cos 	heta) t$.Substituting the values: $x = (\sqrt{800} \cos 45^{\circ}) 	imes 3$.$x = \sqrt{800} 	imes \frac{1}{\sqrt{2}} 	imes 3 = \sqrt{400} 	imes 3 = 20 	imes 3 = 60 \ m$.Thus,the distance travelled in the first $3 \ s$ is $60 \ m$.

In a sports event,a disc is thrown such that it reaches its maximum range of $80 \ m$. The distance travelled in the first $3 \ s$ is $(g = 10 \ m/s^2)$. (in $m$)

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