Two balls are thrown as shown in the figure a | vedclass

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(B) For the first ball thrown vertically upwards with velocity $u_1$,the time of flight is $T_1 = \frac{2u_1}{g}$.For the second ball thrown at an ang

Vedclass · Accepted Answer

$1 : 1$

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(B) For the first ball thrown vertically upwards with velocity $u_1$,the time of flight is $T_1 = \frac{2u_1}{g}$.For the second ball thrown at an angle $(90^\circ - 	heta)$ with the horizontal with velocity $u_2$,the time of flight is $T_2 = \frac{2u_2 \sin(90^\circ - 	heta)}{g} = \frac{2u_2 \cos 	heta}{g}$.Given that $T_1 = T_2$,we have $\frac{2u_1}{g} = \frac{2u_2 \cos 	heta}{g}$,which implies $u_1 = u_2 \cos 	heta$.The maximum height attained by the first ball is $H_1 = \frac{u_1^2}{2g}$.The maximum height attained by the second ball is $H_2 = \frac{(u_2 \sin(90^\circ - 	heta))^2}{2g} = \frac{u_2^2 \cos^2 	heta}{2g}$.Taking the ratio of the heights: $\frac{H_1}{H_2} = \frac{u_1^2 / 2g}{u_2^2 \cos^2 	heta / 2g} = \frac{u_1^2}{u_2^2 \cos^2 	heta}$.Substituting $u_1 = u_2 \cos 	heta$,we get $\frac{H_1}{H_2} = \frac{(u_2 \cos 	heta)^2}{u_2^2 \cos^2 	heta} = 1$.Thus,the ratio is $1 : 1$.

Two balls are thrown as shown in the figure and they reach the ground in the same time. What is the ratio of the maximum heights attained by them?

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