A body of mass m thrown up vertically with ve | vedclass

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Question

(B) For the first body thrown vertically upwards with velocity $v_1$, the time taken to reach maximum height is $t_1 = \frac{v_1}{g}$ and the maximum

Vedclass · Accepted Answer

$4: 1$

Vedclass · Answer

(B) For the first body thrown vertically upwards with velocity $v_1$, the time taken to reach maximum height is $t_1 = \frac{v_1}{g}$ and the maximum height is $h_1 = \frac{v_1^2}{2g}$.For the second body projected with velocity $v_2$ at an angle $	heta$, the time taken to reach maximum height is $t_2 = \frac{v_2 \sin 	heta}{g}$ and the maximum height is $h_2 = \frac{v_2^2 \sin^2 	heta}{2g}$.Given $t_1 = 2t_2$, we have $\frac{v_1}{g} = 2 \left( \frac{v_2 \sin 	heta}{g} ight)$, which simplifies to $v_1 = 2 v_2 \sin 	heta$.Now, the ratio of heights is $\frac{h_1}{h_2} = \frac{v_1^2 / 2g}{v_2^2 \sin^2 	heta / 2g} = \frac{v_1^2}{v_2^2 \sin^2 	heta}$.Substituting $v_1 = 2 v_2 \sin 	heta$, we get $\frac{h_1}{h_2} = \frac{(2 v_2 \sin 	heta)^2}{v_2^2 \sin^2 	heta} = \frac{4 v_2^2 \sin^2 	heta}{v_2^2 \sin^2 	heta} = 4$.Thus, the ratio is $4:1$.

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