(^426) „„„„„ A Textbook of Engineering Mechanics
Fig. 20.6.
(c) Velocity of the second particle for equal time of flight
We know that time of flight of a projectile
2sinu
t
g
α

∴ For equal time of flight
2sinuu112 22 sin
gg
αα

2 × 5 sin 60° = 2u 2 sin 45°
or^2
sin 60 0.866
5 5 6.12 m/s.
sin 45 0.707
u
°
=× =× =
°
Ans
Example 20.10. A particle is projected from the base of a hill whose shape is that of a right
circular cone with axis vertical. The projectile grazes the vertex and strikes the hill again at a point
on the base. If θbe the semi-vertical angle of the cone, h its height, u the initial velocity of the
projectile and αthe angle of projection measured from the horizontal, show that :
tan α = 2 cot θ and
ugh2 tan^221
2
⎛⎞
=+ θ⎜⎟
⎝⎠
.
where g is acceleration due to gravity.
Solution. Given : Semi-vertical angle = θ; Initial velocity of the projectile = u and angle of
projection with the horizontal = α
(i) We know that maximum height through which the particle will
rise,
(^22) sin
2
u
h
g
α

and horizontal range OB
u^2 sin 2
R
g
α

From the geometry of the figure we find that
22
2
sin
2
2 2
cot
sin 2
2
u
hh g
R R u
g
α
×
θ= = =
α
sin^22 sin tan
sin 2 2 sin cos 2
ααα
== =
ααα
∴ tan α = 2 cot θ Ans.
(ii) The above equation, may be written as
11
tan 2 cot

αθ or
1
cot tan
2
α= θ
∴
cot^221 tan
4
α= θ ...(Squaring both sides)
We know that maximum height through which the particle will rise,
(^22) sin
2
u
h
g
α

∴^222
2
2cosec
sin
gh
ugh== α
α