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(C) For maximum horizontal range,the angle of projection is $	heta = 45^\circ$.The maximum range is given by $R_{\max} = \frac{u^2}{g} = 1.6 \; m$.Ta

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$20\sqrt{2} \; m$

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(C) For maximum horizontal range,the angle of projection is $	heta = 45^\circ$.The maximum range is given by $R_{\max} = \frac{u^2}{g} = 1.6 \; m$.Taking $g = 10 \; m/s^2$,we have $u^2 = 1.6 	imes 10 = 16$,so $u = 4 \; m/s$.The horizontal component of velocity is $v_x = u \cos 	heta = 4 \cos 45^\circ = 4 	imes \frac{1}{\sqrt{2}} = 2\sqrt{2} \; m/s$.Since the time spent on the ground is negligible,the grasshopper is effectively in the air for the entire duration of $10 \; s$.The total horizontal distance covered in time $t = 10 \; s$ is $S = v_x 	imes t$.$S = 2\sqrt{2} 	imes 10 = 20\sqrt{2} \; m$.

$A$ grasshopper can jump a maximum distance of $1.6 \; m$. It spends a negligible amount of time on the ground. How far can it go in $10 \; s$?

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