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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 `3.0` km away. By adjusting its angle of projection, can one hope to hit a target 5.0 km away? Assume the muzzle speed to be fixed, and neglect air resistance.

Text Solution

Verified by Experts

We are given that angle of projection with the horizontal, Q = 30°, horizontal range R = 3 km
As `R=(v_(0)^(2)sin 2 theta)/g, 3=(v_(0)^(2) sin 60^(@))/g =v_(0)^(2)/g xx sqrt(3)/2`
or `v_(0)^(2)/g =2sqrt(3)` km
Since the muzzle speed `v_(0)` is fixed,
`R_("max") =v_(0)^(2)/g =2sqrt(3) =2 xx 1.732 = 3.464` km
Obviously, it is not possible to hit the target 5km away.
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