Question

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.

Answer 1

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.