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 away? Assume the muzzle speed to be fixed and neglect air resistance.
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It is possible to hit a target 5.0 Km away by adjusting the angle of projection of the fired bullet.
1. Case 1: when angle of projection is 30 degree.
2. Here range = *sin (2θ)/g.
3. Now range given is 3000 m. So equating we get:
i. 3000 = *sin (2θ)/g.
ii. 3000*g/sin (2θ) =
iii. 3000*9.81/sin(60) =
iv. u= 185 meters per seconds (approx)
4. Now when the range is 5000 m.
i. 5000= *sin (2θ)/g.
ii. sin(2θ) = 5000*9.81/185*185
=1.433
iii. θ= 46 degree (approx)
5. Since this angle is feasible so by adjusting its angle of projection, can one hope to hit a target 5.0 Km away.
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