(^442) „„„„„ A Textbook of Engineering Mechanics
Range of the projectile
We know that range of the projectile,
[]
2
cos^2 sin (2 – ) – sin m
u
R
g
=αββ
β
[]
2
2
(30)
sin (2 55 – 20 ) – sin 20 m
9.8 cos 20
=×°°°
°
2 []
900
sin 90 – sin 20 m
9.8 (0.9397)
=°°
= 104.0 (1 – 0.3420) = 68.43 m Ans.
Time of flight
We also know that the time of flight,
2sin( –) 2 30sin(55–20)
cos 9.8 cos 20
u
t
g
α β ×°°

β°
60 sin 35
9.8 cos 20
×°

°
s
60 0.5736
3.74 s
9.8 0.9397
×

×^
Ans.
Note : Since the angle of projection is for the maximum range, therefore the range may also
be found out from the relation :
(^22) (30) 900
(1 sin ) 9.8 (1 sin 20 ) 9.8 (1 0.3420)
u
R
g
== =
+β + ° +
m
= 68.43 m Ans.
Example 20.25. A plane has a rise of 5 in 12. A shot is projected with a velocity of 200 m/s at
an elevation of 30°. Find the range of the plane, if (a) the shot is fired up the plane, (b) the shot is fired
down the plane.
Solution. Given : tan β = 5/12 = 0.4167 or β = 22.6° ; Velocity of projection with the
horizontal (u) = 200 m/s and angle of projection (α) = 30°.
(a) Range of the plane, when the shot is fired up the plane
We know that range of the plane, when the shot is fired up the plane,
[]
2
1 2 sin (2 – ) – sin
cos
u
R
g
=αββ
β
[]
2
2
(200)
sin (2 30 – 22.6 ) – sin 22.6
9.8 cos 22.6
=×°°°
°
2 []
40 000
sin 37.4 – sin 22.6
9.8 (0.9231)
=°°
= 4790 (0.6072 – 0.3846) = 1066 m Ans.
(b) Range of the plane, when the shot is fired down the plane
We know that range of the plane, when the shot is fired down the plane,
[]
2
2 2 sin (2 ) sin
cos
u
R
g
=α+β + β
β