Engineering Mechanics

Chapter 20 : Projectiles „„„„„ 441

Now for maximum range, substituting the value of αin equation (i),

2

max 2 sin 2 42 – – sin

cos

u

R

g

⎧⎫⎡⎛⎞πβ ⎤

=+⎨⎬⎢⎥⎜⎟ββ

β⎩⎭⎣⎦⎝⎠

2

cos^2 sin^2 – – sin

u

g

⎡ ⎛⎞π ⎤

=+⎢⎥⎜⎟ββ β

β⎣⎦⎝⎠

2

cos^2 sin^2 – sin

u

g

⎡ ⎛⎞π ⎤

= ⎢⎥⎜⎟ β

β⎣⎦⎝⎠

[]

2

cos^2 1–sin

u

g

= β

β

... sin^1

2

⎡ ⎛⎞π ⎤

⎢⎥⎜⎟=

⎣⎦⎝⎠

Q

2

2

(1 – si n )

(1 – si n )

u

g

β

=

β

...(Q sin^2 β + cos^2 β = 1)

2

(1 s i n )

u

g

=

+ β

Notes : 1. When the projectile is projected on a downward inclined plane, then the range of flight

will be given by substituting –βinstead of +βin the above equation. Therefore range of

flight in this case,

[]

2

cos^2 sin (2 ) sin

u

R

g

=α+β+ β

β

2.When the projectile is projected on a downward inclined plane, the range will be maxi-

mum, when

• 22

⎛⎞π β

α=⎜⎟

⎝⎠

3.When the projectile is projected on a downward inclined plane,the value of maximum

range will be

2

max (1 – s i n )

u

R

g

=
β
Example 20.24. A particle is projected from a point, on an inclined plane,with a velocity of
30 m/s. The angle of projection and the angle of plane are 55° and 20° to the horizontal respectively.
Show that the range up the plane is maximum one for the given plane. Find the range and the time of
flight of the particle.

Solution. Given : Velocity of projection (u) = 30 m/s ; Angle of projection with the horizontal

(α) = 55° and angle of plane (β) = 20°

Maximum Range

We know that for maximum range, the angle of projection,

180 20

55

42 4 2

πβ ° °

α= + = + = °

Since the given angle of projection is 55°, therefore range up the plane is maximum one for

the given plane. Ans.