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Question

(A) The time of flight $T$ for a projectile is given by $T = \frac{2u \sin 	heta}{g}$,where $	heta$ is the angle of projection with the horizontal.F

Vedclass · Accepted Answer

$	an \alpha : 1$

Vedclass · Answer

(A) The time of flight $T$ for a projectile is given by $T = \frac{2u \sin 	heta}{g}$,where $	heta$ is the angle of projection with the horizontal.For the first particle,the angle with the horizontal is $	heta_1 = \alpha$. Thus,$T_1 = \frac{2u \sin \alpha}{g}$.For the second particle,the angle with the vertical is $\alpha$,so the angle with the horizontal is $	heta_2 = 90^\circ - \alpha$. Thus,$T_2 = \frac{2u \sin(90^\circ - \alpha)}{g} = \frac{2u \cos \alpha}{g}$.The ratio of their times of flight is $\frac{T_1}{T_2} = \frac{\frac{2u \sin \alpha}{g}}{\frac{2u \cos \alpha}{g}} = \frac{\sin \alpha}{\cos \alpha} = 	an \alpha$.Therefore,the ratio is $	an \alpha : 1$.

$A$ particle is projected with velocity $u$ at an angle $\alpha$ with the horizontal,and another particle is projected with the same velocity $u$ at an angle $\alpha$ with the vertical. What is the ratio of their times of flight?

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