A particle is thrown with a speed u at an ang | vedclass

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

Question

(C) In projectile motion,the horizontal component of velocity remains constant throughout the flight because there is no acceleration in the horizonta

Vedclass · Accepted Answer

$v=u \cos 	heta \sec \phi$

Vedclass · Answer

(C) In projectile motion,the horizontal component of velocity remains constant throughout the flight because there is no acceleration in the horizontal direction.Let the initial velocity be $u$ at an angle $	heta$ with the horizontal. The horizontal component is $u_x = u \cos 	heta$.When the particle is at a point where its velocity is $v$ at an angle $\phi$ with the horizontal,the horizontal component of this velocity is $v_x = v \cos \phi$.Since the horizontal component remains constant,we have $v_x = u_x$.Therefore,$v \cos \phi = u \cos 	heta$.Solving for $v$,we get $v = \frac{u \cos 	heta}{\cos \phi} = u \cos 	heta \sec \phi$.

$A$ particle is thrown with a speed $u$ at an angle $\theta$ with the horizontal. When the particle makes an angle $\phi$ with the horizontal,its speed changes to $v$,where

Similar Questions

What is the range of a projectile? Also,provide the velocity of the projectile at its maximum height.

$A$ cricket ball is hit at $30^{\circ}$ with the horizontal with kinetic energy $K$. The kinetic energy at the highest point is

$A$ ball can be thrown to a maximum horizontal distance of $100\,m$. To what maximum height can it be thrown?

An aeroplane flying at a constant velocity releases a bomb. As the bomb drops down from the aeroplane,

The equation for the trajectory of a projectile is $y = \left( \frac{x}{\sqrt{3}} - \frac{x^2}{60} \right) \text{ m}$. The velocity of projection of the projectile is (Acceleration due to gravity $g = 10 \text{ m s}^{-2}$) (in $\text{ m s}^{-1}$)

Vedclass Test Series

Exam Paper Generator

Online Exam Module