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A bullet fired at an angle of 30^(@) wit...

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.

A

Not possible

B

Possible

C

information insufficient

D

None of these

Text Solution

Verified by Experts

The correct Answer is:
A
Angle of projection `(theta)=30^(@)`
Horizontal range `(R)= 3k, =3000 m`
Horizontal range`(R)=(u^(2) sin 2 theta)/(g)`
`or (u^(2))/(g)=(R)/(sin 2 theta) or (u^(2))/(g)=(3000)/(sin 60^(@))=(3000)/(sqrt(3)//2)`
or `(u^(2))/(g)=(6000)/(sqrt(3))`
When bullet is fired at an angle of projection `45^(@)` then horizontal range maximum.
`:.R_("max")=(u^(2)sin(2xx45^(@)))/(g)=(u^(2))/(g)`
`=(6000)/(sqrt(3))=2000sqrt(3)=3464m`
Therefore, bulet cannot be fired upto 5000m with the same muzzle speed
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