A body is projected horizontally from the top | vedclass

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

Question

(C) In horizontal projectile motion,the horizontal component of velocity $(v_x)$ remains constant throughout the flight.Given,initial horizontal veloc

Vedclass · Accepted Answer

$18$

Vedclass · Answer

(C) In horizontal projectile motion,the horizontal component of velocity $(v_x)$ remains constant throughout the flight.Given,initial horizontal velocity $v_x = 18 \, m s^{-1}$.Let $v_y$ be the vertical component of velocity when the body strikes the ground.The angle of impact $	heta$ with the horizontal is given by $	an 	heta = \frac{v_y}{v_x}$.Given $	heta = 45^{\circ}$,so $	an 45^{\circ} = \frac{v_y}{18}$.Since $	an 45^{\circ} = 1$,we have $1 = \frac{v_y}{18}$.Therefore,$v_y = 18 \, m s^{-1}$.

$A$ body is projected horizontally from the top of a tower with an initial velocity of $18 \, m s^{-1}$. It hits the ground at an angle of $45^{\circ}$. What is the vertical component of velocity when it strikes the ground? $(ms^{-1})$

Similar Questions

At what point of a projectile motion are acceleration and velocity perpendicular to each other?

$A$ projectile is thrown into space so as to have maximum horizontal range $R$. Taking the point of projection as the origin,the coordinates of the point where the speed of the particle is minimum are:

At the top of the trajectory of a projectile,the directions of its velocity and acceleration are

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:

$A$ body is projected at such an angle that the horizontal range is three times the greatest height. The angle of projection is

Vedclass Test Series

Exam Paper Generator

Online Exam Module