Two particles are projected from the same poi | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(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.

Similar Questions

If $\alpha$ is angular acceleration,$\omega$ is angular velocity,and $a$ is the centripetal acceleration,then which of the following is true?

$A$ particle leaves the origin with an initial velocity $\vec{v} = (3 \hat{i}) \text{ m s}^{-1}$ and a constant acceleration $\vec{a} = (-1 \hat{i} - 0.5 \hat{j}) \text{ m s}^{-2}$. The position vector of the particle,when it reaches its maximum $x$-coordinate,is:

$A$ particle moves in the $xy$-plane with velocity $v_x = 8t - 2$ and $v_y = 2$. If it passes through the point $(14, 4)$ at $t = 2 \, s$,the equation of its path is:

$A$ particle moves towards east with velocity $5\, m/s$. After $10\, s$,its direction changes towards north with the same velocity. The average acceleration of the particle is

Two bodies were thrown simultaneously from the origin: one straight up and the other at an angle of $60^{\circ}$ to the vertical. The initial velocity of each body is $10 \ m \ s^{-1}$. Neglecting air resistance,the distance between the two bodies after $t=2 \ s$ is (Use $g=10 \ m \ s^{-2}$):

Vedclass Test Series

Exam Paper Generator

Online Exam Module