Question

A bullet fired at an angle of 30° with the horizontal hits the ground 3.0 km away. By adjusting its angle of projection, can one hope to hit a target 5.0 km away? Assume the muzzle speed to the fixed, and neglect air resistance.

Answer 1

No

Range, R = 3 km

Angle of projection, θ = 30°

Acceleration due to gravity, g = 9.8 m/s²

Horizontal range for the projection velocity u₀, is given by the relation:

R = `(u^2_0sin2theta)/g`

`3 = u_0^2/g sin 60^@`

`u_0^2/g =2sqrt3`  ..(i)

The maximum range (R_max) is achieved by the bullet when it is fired at an angle of 45° with the horizontal, that is,

`R_"max" = (u_0^2)/g`  ..(ii)

On comparing equations (i) and (ii), we get:

`R_"max" = 3sqrt3 = 2 xx 1.732 = 3.46 km`

Hence, the bullet will not hit a target 5 km away

Answer 2

Here R = 3 km = 3000 m, `theta = 30^@`, g = 9.8 `ms^(-2)`

As R = `(u^2sin 2theta)/g`

`=> 3000 = (u^2sin 2 xx 30^@)/9.8 = (u^2sin 60)/9.8`

`=> u^2 = (3000xx 9.8)/(sqrt3"/2") = 3464 xx 9.8`

Also `R' = (u^2sin 2theta')/g => 5000 = (3464 xx 9.8 xx sin 2theta)/9.8`

i.e `sin 2theta' = 5000/3464 = 1.44`

Which is impossible because sine of an angle cannot be more than 1. Thus this target cannot be hoped to be hit.