A particle is projected at an angle of 60^{ci | vedclass

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

Question

(D) Initial velocity $u = 10 \sqrt{3} \ m/s$ at $	heta = 60^{\circ}$.Horizontal component: $u_x = u \cos 60^{\circ} = 10 \sqrt{3} 	imes (1/2) = 5 \s

Vedclass · Accepted Answer

$90$

Vedclass · Answer

(D) Initial velocity $u = 10 \sqrt{3} \ m/s$ at $	heta = 60^{\circ}$.Horizontal component: $u_x = u \cos 60^{\circ} = 10 \sqrt{3} 	imes (1/2) = 5 \sqrt{3} \ m/s$.Vertical component: $u_y = u \sin 60^{\circ} = 10 \sqrt{3} 	imes (\sqrt{3}/2) = 15 \ m/s$.Initial velocity vector: $\vec{v}_i = 5 \sqrt{3} \hat{i} + 15 \hat{j}$.After $t = 2 \ s$,horizontal velocity $v_x = u_x = 5 \sqrt{3} \ m/s$.Vertical velocity $v_y = u_y - gt = 15 - (10 	imes 2) = 15 - 20 = -5 \ m/s$.Final velocity vector: $\vec{v}_f = 5 \sqrt{3} \hat{i} - 5 \hat{j}$.Calculate the dot product: $\vec{v}_i \cdot \vec{v}_f = (5 \sqrt{3})(5 \sqrt{3}) + (15)(-5) = 75 - 75 = 0$.Since the dot product is $0$,the angle between the vectors is $90^{\circ}$.

$A$ particle is projected at an angle of $60^{\circ}$ with the horizontal from the ground with a velocity $10 \sqrt{3} \ m/s$. The angle between the velocity vector after $2 \ s$ and the initial velocity vector is $(g = 10 \ m/s^2)$. (in $^{\circ}$)

Similar Questions

Two projectiles thrown at $30^{\circ}$ and $45^{\circ}$ with the horizontal respectively,reach the maximum height in the same time. The ratio of their initial velocities is

If $x$ and $y$ coordinates of a projectile as a function of time are given as $24t$ and $43.6t - 4.9t^2$, respectively, then the angle (in degrees) made by the projectile with horizontal when $t = 2 \text{ s}$ is . . . . . . .

The trajectory of a projectile projected from the origin is given by the equation $y = x - \frac{2x^2}{5}$. The initial velocity of the projectile is:

If we can throw a ball up to a maximum height $H$,the maximum horizontal distance to which we can throw it is

$A$ boy throws a ball upwards with velocity $v_0 = 20\, m/s$. The wind imparts a horizontal acceleration of $4\, m/s^2$ to the ball. The angle $\theta$ from the vertical at which the ball must be thrown so that the ball returns to the boy's hand is $(g = 10\, m/s^2)$

Vedclass Test Series

Exam Paper Generator

Online Exam Module