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(D) Let the angles with the horizontal be $	heta_A$ and $	heta_B$. Given angles with vertical are $45^{\circ}$ and $60^{\circ}$,so $	heta_A = 90^{\

Vedclass · Accepted Answer

$1 : \sqrt{2}$

Vedclass · Answer

(D) Let the angles with the horizontal be $	heta_A$ and $	heta_B$. Given angles with vertical are $45^{\circ}$ and $60^{\circ}$,so $	heta_A = 90^{\circ} - 45^{\circ} = 45^{\circ}$ and $	heta_B = 90^{\circ} - 60^{\circ} = 30^{\circ}$.For a projectile thrown from height $h$,the time of flight $T$ is given by $h = -u \sin 	heta T + \frac{1}{2} g T^2$. Since $T$ is the same for both,the vertical components of initial velocity must be equal: $v_A \sin 	heta_A = v_B \sin 	heta_B$.$v_A \sin 45^{\circ} = v_B \sin 30^{\circ} \implies v_A (\frac{1}{\sqrt{2}}) = v_B (\frac{1}{2}) \implies \frac{v_A}{v_B} = \frac{\sqrt{2}}{2} = \frac{1}{\sqrt{2}}$.Also,the range $R = (v \cos 	heta) T$. Since $R$ and $T$ are same,$v_A \cos 	heta_A = v_B \cos 	heta_B$.$v_A \cos 45^{\circ} = v_B \cos 30^{\circ} \implies v_A (\frac{1}{\sqrt{2}}) = v_B (\frac{\sqrt{3}}{2}) \implies \frac{v_A}{v_B} = \frac{\sqrt{6}}{2} = \sqrt{\frac{3}{2}}$.Since the two conditions lead to different ratios,the problem statement is physically inconsistent. However,if we assume the vertical components are equal to satisfy the time of flight condition,the ratio is $1 : \sqrt{2}$.

Projectiles $A$ and $B$ are thrown at angles of $45^{\circ}$ and $60^{\circ}$ with the vertical,respectively,from the top of a $400 \ m$ high tower. If their ranges and times of flight are the same,the ratio of their speeds of projection $v_A : v_B$ is:

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