1. At an angle of 30° from horizontal a bullet is
fired from a gun, which falls 3 km away on
the ground. Neglecting the air friction, show
if an object 9 km away can be targeted?
Answers
Answer:
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=
g
u
o
2
Sin 60
0
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
Answer:
Range, R=3km
Angle of projection, =30°
Acceleration due to gravity, g=9.8m/s2
Horizontal range for the projection velocity u
, is given by the relation:
R = u2 sin 2θ / g
3 = u2 sin 60° / g
u2 / g = 2√3 ................ (1)
The maximum range (Rmax) is achieved by the bullet when it is fired at an angle of 45 with the horizontal, that is,
Rmax = u2/ g ....(ii)
On comparing equations (i) and (ii),
we get:
Rmax = 2×1.732
=3.46 km
hope it will help you....