Engineering Mechanics

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(^432) „„„„„ A Textbook of Engineering Mechanics
Fig. 20.12.
Time when the shots meet after they leave the guns
Let the two shots meet at C as shown in Fig. 20.12.
Now let t= Time in seconds, when the two shots meet after they leave the
guns.
x= Horizontal distance between A and C
y= Vertical distance between A and C
∴ Horizontal distance between A and B (i.e.^ AD)
30 3
30 cos 30 15 3 m
2
=°==
We know that distance covered by the shot A in t seconds,
x = Horizontal component of vA × t
3
350 cos 30 350 175 3
2
=°×=×=ttt ...(i)
Similarly, distance covered by the shot B in t seconds
3
(15 3 – ) 300 cos 30 300 150 3
2
xttt= °×= ×= ...(ii)
Adding equation (i) and (ii),
15 3 175 3=×+×tt150 3
or 15 = 175 t + 150 t = 325 t

15
0.046 s
325
t== Ans.
Point where the two shots meet
Substituting this value of t in equation (i),
x=×=175 3 0.046 13.94 m Ans.
We know that vertical component of vA
= 350 sin 30° = 300 × 0.5 = 175 m/s
∴Vertical distance between A and C


-^1122 (175 0.046) – 9.8 (0.046) m
22


yut gt==× ×⎛⎞⎜⎟
⎝⎠
= 8.04 m Ans.
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