A ball of mass m is thrown vertically upward. | vedclass

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(A) Let the initial velocity of the first ball be $u_1$ and the second ball be $u_2$.For the first ball thrown vertically upward,the time of flight is

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$1$

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(A) Let the initial velocity of the first ball be $u_1$ and the second ball be $u_2$.For the first ball thrown vertically upward,the time of flight is $T_1 = \frac{2u_1}{g}$.For the second ball thrown at an angle $	heta$ with the vertical,the vertical component of velocity is $u_{2y} = u_2 \cos 	heta$. The time of flight is $T_2 = \frac{2u_2 \cos 	heta}{g}$.Given $T_1 = T_2$,we have $u_1 = u_2 \cos 	heta$.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 $H_2 = \frac{(u_2 \cos 	heta)^2}{2g} = \frac{u_2^2 \cos^2 	heta}{2g}$.Since $u_1 = u_2 \cos 	heta$,it follows that $H_1 = H_2$.The ratio of the heights is $\frac{H_1}{H_2} = 1 = \frac{1}{x}$.Therefore,$x = 1$.

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