(^436) „„„„„ A Textbook of Engineering Mechanics
Example 20.20. A particle is projected upwards with a velocity of 100 m/s at an angle of
30 ° with the horizontal.
Find the time, when the particle will move perpendicular to its initial direction.
Solution. Given : Initial velocity of projection (u) = 100 m/s and angle of projection
with the horizontal (α) = 30°
Fig. 20.15.
Let t= Time from the instant of projection, when the particle will
move perpendicular to its initial direction.
We know that when the particle will move perpendicular to its initial direction, it will make an
angle of 90° – 30° = 60° with the horizontal, but in the downward direction as shown in Fig. 20.14.
Therefore actual angle,
θ= (– 60°) ...(Minus sign due to downward)
We also know that the angle, which the particle makes with the horizontal after t seconds (θ),
sin – 100 sin 30 – 9.8
tan (– 60 )
cos 100 cos 30
ugt t
u
α°
°= =
α°
(100 0.5) – 9.8 50 – 9.8
–1.732
100 0.866 86.6
× tt

×

• 150 = 50 – 9.8 t

∴

50 150

20.4 s

9.8

t

+

== Ans.

20.10.VELOCITY AND DIRECTION OF MOTION OF A PROJECTILE, AT A

GIVEN HEIGHT ABOVE THE POINT OF PROJECTION

Consider a projectile projected from O as shown in Fig. 20.16.

Let u= Initial velocity of projection, and

α= Angle of projection with the

horizontal.

After reaching a height h, let the projectile reach at any point P

as shown in Fig. 20.15.

Let v = Velocity of the projectile at P, and

θ = Angle, which the projectile at P

makes with the horizontal. Fig. 20.16.