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(D) Given: $m = 1 \, kg$,$u = \sqrt{20} \, m/s$,$R = \sqrt{3} \, m$,$g = 10 \, m/s^2$. Using range formula: $R = \frac{u^2 \sin 2	heta}{g} \Rightarro

Vedclass · Accepted Answer

maximum potential energy during its motion can be $6 \, J$.

Vedclass · Answer

(D) Given: $m = 1 \, kg$,$u = \sqrt{20} \, m/s$,$R = \sqrt{3} \, m$,$g = 10 \, m/s^2$. Using range formula: $R = \frac{u^2 \sin 2	heta}{g} \Rightarrow \sqrt{3} = \frac{20 \sin 2	heta}{10} \Rightarrow \sin 2	heta = \frac{\sqrt{3}}{2}$. Thus,$2	heta = 60^\circ$ or $120^\circ$,so $	heta = 30^\circ$ or $60^\circ$. For $	heta = 30^\circ$: $H = \frac{u^2 \sin^2 30^\circ}{2g} = \frac{20 	imes (1/4)}{20} = 0.25 \, m$. For $	heta = 60^\circ$: $H = \frac{u^2 \sin^2 60^\circ}{2g} = \frac{20 	imes (3/4)}{20} = 0.75 \, m$. Minimum velocity $v_{\min} = u \cos 	heta$. For $	heta = 30^\circ$,$v_{\min} = \sqrt{20} 	imes \frac{\sqrt{3}}{2} = \sqrt{15} \, m/s$. Time of flight $T = \frac{2u \sin 	heta}{g}$. For $	heta = 60^\circ$,$T = \frac{2 \sqrt{20} 	imes \frac{\sqrt{3}}{2}}{10} = \sqrt{\frac{3}{5}} \, s$. Maximum potential energy $U_{\max} = mgH_{\max} = 1 	imes 10 	imes 0.75 = 7.5 \, J$. Comparing with options: $(a)$ is correct ($0.25 \, m$ is possible),$(b)$ is correct ($\sqrt{15} \, m/s$ is possible),$(c)$ is correct ($\sqrt{3/5} \, s$ is possible),$(d)$ is incorrect $(7.5 \, J 
eq 6 \, J)$. Thus,option $(d)$ is the incorrect statement.

$A$ projectile of mass $1 \, kg$ is projected with a velocity of $\sqrt{20} \, m/s$ such that it strikes on the same level as the point of projection at a distance of $\sqrt{3} \, m$. Which of the following options are incorrect:

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