Projectile Motion

by • 04/07/2012 • GeneralComments (0)602

Motion in two Dimension

Projectile Motion

A body moving horizontally as well as vertically under the action of gravity simultaneously is called a projectile. The motion of projectile is called projectile motion. The path followed by a projectile is called its trajectory.

Examples of projectile motion are

1. Kicked or thrown balls

2. Jumping animals

3. A bomb released from a bomber plane

4. A shell of a gun.
Analysis of Projectile Motion

Let us consider a body of mass m, projected an angle θ with the horizontal with a velocity V0. We made the following three assumptions.

1. The value of g remains constant throughout the motion.

2. The effect of air resistance is negligible.

3. The rotation of earth does not affect the motion.
Horizontal Motion

Acceleration : ax = 0

Velocity : Vx = Vox

Displacement : X = Vox t
Vertical Motion

Acceleration : ay = – g

Velocity : Vy = Voy – gt

Displacement : Y = Voy t – 1/2 gt2
Initial Horizontal Velocity

Vox = Vo cos θ …………………. (1)
Initial Vertical Velocity

Voy = Vo sin θ …………………. (2)

Net force W is acting on the body in downward vertical direction, therefore, vertical velocity continuously changes due to the acceleration g produced by the weight W.

There is no net force acting on the projectile in horizontal direction, therefore, its horizontal velocity remains constant throughout the motion.
X – Component of Velocity at Time t (Vx)

Vx = Vox = Vo cos θ ……………….. (3)
Y – Component of Velocity at Time t (Vy)

Data for vertical motion

Vi = Voy = Vo sin θ

a = ay = – g

t = t

Vf = Vy = ?

Using Vf = Vi + at

Vy = Vo sin θ – gt ……………….. (4)
Range of the Projectile (R)

The total distance covered by the projectile in horizontal direction (X-axis) is called is range

Let T be the time of flight of the projectile.

Therefore,

R = Vox x T ………….. {since S = Vt}

T = 2 (time taken by the projectile to reach the highest point)

T = 2 Vo sin θ / g

Vox = Vo cos θ

Therefore,

R = Vo cos θ x 2 Vo sin θ / g

R = Vo2 (2 sin θ cos θ) / g

R = Vo2 sin 2 θ / g ……………… { since 2 sin θ cos θ = sin2 θ}

Thus the range of the projectile depends on

(a) The square of the initial velocity

(b) Sine of twice the projection angle θ.

The Maximum Range

For a given value of Vo, range will be maximum when sin2 θ in R = Vo2 sin2 θ / g has maximum value. Since

0 ≤ sin2 θ ≤ 1

Hence maximum value of sin2 θ is 1.

Sin2 θ = 1

2θ = sin(-1) (1)

2θ = 90º

θ = 45º

Therefore,

R(max) = Vo2 / g ; at θ = 45º

Hence the projectile must be launched at an angle of 45º with the horizontal to attain maximum range.
Projectile Trajectory

The path followed by a projectile is referred as its trajectory.

We known that

S = Vit + 1/2 at2

For vertical motion

S = Y

a = – g

Vi = Voy = Vo sin θ

Therefore,

Y = Vo sinθ t – 1/2 g t2 ………………….. (1)

Also

X = Vox t

X = Vo cosθ t ………… { since Vox = Vo cosθ}

t = X / Vo cos θ
(1) => Y = Vo sinθ (X / Vo cos θ) – 1/2 g (X / Vo cos θ)2

Y = X tan θ – gX2 / 2Vo2 cos2 θ

For a given value of Vo and θ, the quantities tanθ, cosθ, and g are constant, therefore, put

a = tan θ

b = g / Vo2 cos2θ

Therefore

Y = a X – 1/2 b X2

Which shows that trajectory is parabola.

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