A bullet fired at an angle of 30^(@) with the horizontal hits the ground 3 km away. By adjusting the angle of projection, can one hope to hit a target 5 km away ? Assume the muzzle speed to be fixed and neglect air resistance.
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Answer:
Range, R=3km
Angle of projection, =30
o
Acceleration due to gravity, g=9.8m/s
2
Horizontal range for the projection velocity u
0
, is given by the relation:
R=
g
u
o
2
Sin 2θ
$$3 = \dfrac{u_o ^2 \ Sin \ 60^0 }{ g}$$
g
u
o
2
=2
3
.......(i)
The maximum range (Rmax) is achieved by the bullet when it is fired at an angle of 45 with the horizontal, that is, Rmax = u
o
2
/ g ....(ii)
On comparing equations (i) and (ii), we get:
R
max
=2×1.732=3.46 km
Hence, the bullet will not hit a target 5 km away.
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