At t = 0,a projectile is fired from a point O | vedclass

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

Question

(A) Given: Initial speed $u = 50 \, m/s$,angle $	heta = 53^{\circ}$,time $t = 2 \, s$.The horizontal and vertical components of initial velocity are:

Vedclass · Accepted Answer

$60\sqrt{2}$

Vedclass · Answer

(A) Given: Initial speed $u = 50 \, m/s$,angle $	heta = 53^{\circ}$,time $t = 2 \, s$.The horizontal and vertical components of initial velocity are:$u_x = u \cos 53^{\circ} = 50 	imes \frac{3}{5} = 30 \, m/s$$u_y = u \sin 53^{\circ} = 50 	imes \frac{4}{5} = 40 \, m/s$At any time $t$,the position coordinates $(x, y)$ of the projectile are given by:$x = u_x t = 30 	imes 2 = 60 \, m$$y = u_y t - \frac{1}{2} g t^2 = 40 	imes 2 - \frac{1}{2} 	imes 10 	imes (2)^2 = 80 - 20 = 60 \, m$The distance $d$ of the particle from the origin $(0, 0)$ at $t = 2 \, s$ is:$d = \sqrt{x^2 + y^2} = \sqrt{60^2 + 60^2} = \sqrt{2 	imes 60^2} = 60\sqrt{2} \, m$.

Similar Questions

$A$ projectile is thrown upward at an angle $60^{\circ}$ with the horizontal. The speed of the projectile is $20 \text{ m/s}$ when its direction of motion is $45^{\circ}$ with the horizontal. The initial speed of the projectile is . . . . . . in $\text{m/s}$.

$A$ man standing on the roof of a house of height $h$ throws one particle vertically downwards and another particle horizontally with the same velocity $u$. The ratio of their velocities when they reach the earth's surface will be

$A$ projectile is thrown into air with velocity $15 \ m/s$ at an angle $30^{\circ}$ with the horizontal. After what time is its direction of motion perpendicular to its initial direction (in $s$)? (Assume $g = 10 \ m/s^2$)

The velocity of a body at time $t = 0$ is $10\sqrt{2} \, m/s$ in the north-east direction and it is moving with an acceleration of $2 \, m/s^2$ directed towards the south. The magnitude and direction of the velocity of the body after $5 \, s$ will be

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