If a → + b → = a → - b → prove that the angle between a → a n d b → i s 90 ° .

We have; a → + b → = a → - b → Squaring both the sides, we get ⇒ a → + b → 2 = a → - b → 2 ⇒ a → + b → . a → + b → = a → - b → . a → - b → . . . . . ∵ A → 2 = A → . A → ⇒ a → . a → + a → . b → + b → . a → + b → . b → = a → . a → - a → . b → - b → . a → + b → . b → ⇒ a → . b → + a → . b → = - a → . b → - a → . b → ⇒ 4 × a → . b → = 0 ⇒ 4 × a b cos θ = 0 ⇒ cos θ = 0 ⇒ θ = cos - 1 ( 0 ) ⇒ θ = 90 °

Show that the maximum height and range of a projectile are U 2 sin 2 θ 2 g a n d U 2 sin 2 θ g respectively where the terms have their regular meanings.

Let a body is projected with an initial velocity ' u ' and with an angle θ to the horizontal. Initial velocity along x - d i r e c t i o n , u x = u cos θ Initial velocity along y - d i r e c t i o n , u y = u sin θ Horizontal Range: It is the distance covered by projectile along the horizontal between the point for projection to the point on ground, where the projectile returns again. It is denoted by R . The horizontal distance covered by the projectile in the to time of flight is called horizontal range. Therefore, R = u cos θ × t . R = u cos θ × 2 u sin θ g = u 2 2 sin θ cos θ g R = u 2 sin 2 θ g . . . . . ∵ 2 sin θ cos θ = sin 2 θ Angle of projection for maximum range: For a given velocity of projection, the horizontal range will be maximum, when sin 2 θ = 1 . ∴ A n g l e o f p r o j e c t i o n f o r m a x i m u m r a n g e i s 2 θ = 90 ° o r θ = 45 ° ∴ R m a x = u 2 g . Maximum height: The vertical distance covered by the projectile until its vertical component becomes zero. Applying; v 2 - u 2 = 2 a s We have, 0 - u 2 sin 2 θ = 2 - g H Maximum height reached, H = u 2 sin 2 θ 2 g

E M B I B E

Intermediate First Year Physics>Chapter 4 - Motion in a Plane>EXERCISES>Q 2

EASY

11th Telangana Board

IMPORTANT

Earn 100

What is relative motion. Explain it?

Important Questions on Motion in a Plane

EASY

11th Telangana Board

IMPORTANT

Intermediate First Year Physics>Chapter 4 - Motion in a Plane>EXERCISES>Q 3

Show that a boat must move at an angle with respect to river water in order to cross the river in minimum time?

EASY

11th Telangana Board

IMPORTANT

Intermediate First Year Physics>Chapter 4 - Motion in a Plane>EXERCISES>Q 4

Define unit vector, null vector and position vector.

EASY

11th Telangana Board

IMPORTANT

Intermediate First Year Physics>Chapter 4 - Motion in a Plane>EXERCISES>Q 5

If

|\vec{a} + \vec{b}| = |\vec{a} - \vec{b}|

prove that the angle between

\vec{a} a n d \vec{b} i s 90 ° .

EASY

11th Telangana Board

IMPORTANT

Intermediate First Year Physics>Chapter 4 - Motion in a Plane>EXERCISES>Q 6

Show that the trajectory of an object thrown at certain angle with the horizontal is a parabola.

EASY

11th Telangana Board

IMPORTANT

Intermediate First Year Physics>Chapter 4 - Motion in a Plane>EXERCISES>Q 7

Explain the terms the average velocity and instantaneous velocity. When are they equal?

EASY

11th Telangana Board

IMPORTANT

Intermediate First Year Physics>Chapter 4 - Motion in a Plane>EXERCISES>Q 8

Show that the maximum height and range of a projectile are

\frac{U^{2} \sin^{2} θ}{2 g} a n d \frac{U^{2} \sin 2 θ}{g}

respectively where the terms have their regular meanings.

EASY

11th Telangana Board

IMPORTANT

Intermediate First Year Physics>Chapter 4 - Motion in a Plane>EXERCISES>Q 9

If the trajectory of a body is parabolic in one reference frame, can it be parabolic in another reference frame that moves at constant velocity with respect to the first reference frame? If the trajectory can be other than parabolic, what else can it be?

EASY

11th Telangana Board

IMPORTANT

Intermediate First Year Physics>Chapter 4 - Motion in a Plane>EXERCISES>Q 10

A force