Engineering Mechanics

Chapter 20 : Projectiles „„„„„ 439

Let u= Initial velocity of projection,

α= Angle of projection with the horizontal.

β= Inclination of the plane OA with the horizontal,

R= Range of flight from O to B, and

t= Time of flight from O to B.

∴ Component of initial velocity, normal to the plane OA

= u sin (α – β) ...(i)

We know that acceleration due to gravity normal to the plane OA

= g cos β ...(ii)

and acceleration due to gravity along the plane OA

= g sin β ...(iii)
Now consider the motion of the projectile normal to the plane. We know that distance covered
by the projectile normal to the plane OA is zero. Therefore substituting these values in the general
equation of motion, i.e.

-^12
2

sut= gt

0sin(–)–(cos)^12

2

=αutgtββ

or

1

0sin(–)–(cos)

2

=αugtββ ... (Dividing both sides by t)

∴

2sin( –)

cos

u

t

g

α β

=

β

Note : When the projectile is projected on a downward inclined plane, the time of flight may

be found out by substituting –β instead of +βin the above equation. Therefore time of flight in this

case,

2sin( )

cos

u

t

g

α+β

=

β

... [Q cos (– β) = cos β]

Example 20.23. A ball is projected from a point with a velocity of 10 m/s on an inclined
plane. the angle of projection and inclination of the plane are 35° and 15° respectively with the
horizontal. Find the time of flight of the ball, when it is projected upwards and downwards the plane.

Solution. Given : Velocity of projection (u) = 10 m/s ; Angle of projection with the horizontal

(α) = 35° and inclination of the plane (β) = 15°

Time of flight when the ball is projected upwards

We know that time of flight when the ball is projected upwards,

1

2 sin ( – ) 2 10 sin (35 – 15 )

cos 9.8 cos 15

u

t

g

α β ×°°

==

β°

20 sin 20

9.8 cos15

°

=

°

s

20 0.342

0.72 s.

9.8 0.9659

×

==

×

Ans