Chapter 20 : Projectiles „„„„„ 425

Solution. Given : Horizontal range (R) = 3 H (where H is the greatest height). ...(i)

Let α = Angle of projection.

We know that horizontal range,

u^2 sin 2

R

g

α

=

and the greatest height

(^22) sin
2
u
H
g
α

Substituting these values of R and H in the given equation (i),
(^222) sin 2 sin
3
2
uu
gg
αα
=×
(^222) 2sin cos sin
3
2
uu
gg
×αα α
=× ...(Q 22sincosα= α α)
or 2 cosα=1.5 sinα
∴
2
tan 1.333
1.5
α= = or α = 53.1° Ans.
Example 20.9. A particle is thrown with a velocity of 5 m/s at an elevation of 60° to the
horizontal. Find the velocity of another particle thrown at an elevation of 45° which will have (a)
equal horizontal range, (b) equal maximum height, and (c) equal time of flight.
Solution. Given : Velocity of projection of first particle (u 1 ) = 5 m/s ; Angle of projection
of first particle with the horizontal (α 1 ) = 60° and angle of projection of second particle with the
horizontal (α 2 ) = 45°
Let u 2 = Velocity of projection of the second particle.
(a) Velocity of the second particle for equal horizontal range
We know that horizontal range of a projectile,
u^2 sin 2
R
g
α

∴ For equal horizontal range
22
uu 1122 sin 2 sin 2
gg
αα

(5)^2 sin (2 × 60°) = u 22 sin (2 × 45°)
or
2
2
sin 120 0.866
25 25 21.65
sin 90 1.0
u
°
=× =× =
°
∴ u 2 = 4.65 m/s Ans.
(b) Velocity of the second particle for equal maximum height
We know that maximum height of a projectile,
(^22) sin
2
u
H
g
α

∴ For equal maximum height
22 22
1122 sin sin
22
uu
gg
αα

(5)^2 sin^2 60° = u 22 sin^2 45°
or
22
2
(^222)
sin 60 (0.866)
25 25 37.5
sin 45 (0.707)
u
°
=× =× =
°
∴ u 2 = 6.12 m/s Ans.