Two particles are projected from the same poi | vedclass

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(D) For the same range with the same speed,the angles of projection must be complementary,so $	heta _1 + 	heta _2 = 90^\circ$,which implies $	heta

Vedclass · Accepted Answer

All of the above

Vedclass · Answer

(D) For the same range with the same speed,the angles of projection must be complementary,so $	heta _1 + 	heta _2 = 90^\circ$,which implies $	heta _1 = 90^\circ - 	heta _2$. This confirms option $A$.The time of flight for a projectile is given by $t = \frac{2u \sin 	heta}{g}$.Thus,$t_1 = \frac{2u \sin 	heta _1}{g}$ and $t_2 = \frac{2u \sin 	heta _2}{g}$.From these,$\frac{t_1}{\sin 	heta _1} = \frac{2u}{g}$ and $\frac{t_2}{\sin 	heta _2} = \frac{2u}{g}$.Therefore,$\frac{t_1}{\sin 	heta _1} = \frac{t_2}{\sin 	heta _2}$,which confirms option $B$.Since $	heta _2 = 90^\circ - 	heta _1$,we have $\sin 	heta _2 = \sin(90^\circ - 	heta _1) = \cos 	heta _1$.Then,$\frac{t_1}{t_2} = \frac{\sin 	heta _1}{\sin 	heta _2} = \frac{\sin 	heta _1}{\cos 	heta _1} = 	an 	heta _1$,which confirms option $C$.Since all options are correct,the correct answer is $D$.

Two particles are projected from the same point with the same speed at different angles $\theta _1$ and $\theta _2$ to the horizontal. They have the same range. Their times of flight are $t_1$ and $t_2$ respectively.

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