B and c only 12 A particle is projected from the origin with speed V m/s at an angle a to theyAhorizontal.a Assuming that the coordinates of the particle at time t are(Vt cos a, Vt sin a – ¿gt2), prove that the horizontal range R of the particlev² sin 2aisX26.b Hence prove that the path of the particle has equation y = xXtan a.R1 -c Suppose that a = 45° and that the particle passes through two points 6metres apart and 4 metres above the point of projection, as shown in the diagram. Let x, and x2 be thex-coordinates of the two points.i Show thatii Use the identity (x2x2 are the roots of the equation x2x1)? = (x2 + x1)²X 1andRx + 4R = 0.4x2x1 to find R.-

Name: B and c only 12 A particle is projected from the origin with speed V m
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Description: B and c only 12 A particle is projected from the origin with speed V m/s at an angle a to theyAhorizontal.a Assuming that the coordinates of the particle at time t are(Vt cos a, Vt sin a – ¿gt2), prove that the horizontal range R of the particlev² si

Given that: Coordinate at any time t: (Vtcosα, Vtsinα-12gt2)

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A particle is projected from the origin with speed V m/s at an angle a to the norizontal. 4 a Assuming that the coordinates of the particle at time t are (Vt cos a, Vt sin a – gt2), prove that the horizontal range R of the particle v² sin 2a is *(1 - Jun 6 b Hence prove that the path of the particle has equation y = x tan a. R c Suppose that a = 45° and that the particle passes through two points 6 metres apart and 4 metres above the point of projection, as shown in the diagram. Let x-coordinates of the two points. and x2 be the i Show that x1 and x2 are the roots of the equation x2 ii Use the identity (x2 - x1)? = (x2 + x1) - 4xx¡ to find R. Rx + 4R = 0.

A particle is projected from the origin with speed V m/s at an angle a to the norizontal. 4 a Assuming that the coordinates of the particle at time t are (Vt cos a, Vt sin a – gt2), prove that the horizontal range R of the particle v² sin 2a is *(1 - Jun 6 b Hence prove that the path of the particle has equation y = x tan a. R c Suppose that a = 45° and that the particle passes through two points 6 metres apart and 4 metres above the point of projection, as shown in the diagram. Let x-coordinates of the two points. and x2 be the i Show that x1 and x2 are the roots of the equation x2 ii Use the identity (x2 - x1)? = (x2 + x1) - 4xx¡ to find R. Rx + 4R = 0.

College Physics

11th Edition

ISBN:9781305952300

Author:Raymond A. Serway, Chris Vuille

Publisher:Raymond A. Serway, Chris Vuille

Chapter1: Units, Trigonometry. And Vectors

Section: Chapter Questions

Problem 1CQ: Estimate the order of magnitude of the length, in meters, of each of the following; (a) a mouse, (b)...

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Concept explainers

Displacement, Velocity and Acceleration

In classical mechanics, kinematics deals with the motion of a particle. It deals only with the position, velocity, acceleration, and displacement of a particle. It has no concern about the source of motion.

Linear Displacement

The term "displacement" refers to when something shifts away from its original "location," and "linear" refers to a straight line. As a result, “Linear Displacement” can be described as the movement of an object in a straight line along a single axis, for example, from side to side or up and down. Non-contact sensors such as LVDTs and other linear location sensors can calculate linear displacement. Non-contact sensors such as LVDTs and other linear location sensors can calculate linear displacement. Linear displacement is usually measured in millimeters or inches and may be positive or negative.

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Question

B and c only

12 A particle is projected from the origin with speed V m/s at an angle a to the
yA
horizontal.
a Assuming that the coordinates of the particle at time t are
(Vt cos a, Vt sin a – ¿gt2), prove that the horizontal range R of the particle
v² sin 2a
is
X2
6.
b Hence prove that the path of the particle has equation y = x
X
tan a.
R
1 -
c Suppose that a = 45° and that the particle passes through two points 6
metres apart and 4 metres above the point of projection, as shown in the diagram. Let x, and x2 be the
x-coordinates of the two points.
i Show that
ii Use the identity (x2
x2 are the roots of the equation x2
x1)? = (x2 + x1)²
X 1
and
Rx + 4R = 0.
4x2x1 to find R.
-

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