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प्रश्न
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.
उत्तर १
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 (Rmax) 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
उत्तर २
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.
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