A bullet is fired from a height of 120m at a velocity of 360 km per hour at an angle of 30 degree upward. Find - (1) horizontal range of bullet . (2) maximum height reached by Bullet. (3) final velocity. (4) position of bullet at t = 6 seconds. It is the problem of projectile motion.

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Arun · Answer

Dear Ram Horizontal Range = u² * sin2@/g Maximum height = u²* sin²@ /2g Final velocity will be the same as initial velocity but components get changed. Put the values and get the answers. Regards Arun (askIITians forum expert) If any query arise, please feel free to ask

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A bullet is fired from a height of 120m at a velocity of 360 km per hour at an angle of 30 degree upward. Find - (1) horizontal range of bullet . (2) maximum height reached by Bullet. (3) final velocity. (4) position of bullet at t = 6 seconds. It is the problem of projectile motion.

A bullet is fired from a height of 120m at a velocity of 360 km per hour at an angle of 30 degree upward. Find - (1) horizontal range of bullet . (2) maximum height reached by Bullet. (3) final velocity. (4) position of bullet at t = 6 seconds. It is the problem of projectile motion.

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A bullet is fired from a height of 120m at a velocity of 360 km per hour at an angle of 30 degree upward. Find - (1) horizontal range of bullet . (2) maximum height reached by Bullet. (3) final velocity. (4) position of bullet at t = 6 seconds. It is the problem of projectile motion.

A bullet is fired from a height of 120m at a velocity of 360 km per hour at an angle of 30 degree upward. Find - (1) horizontal range of bullet . (2) maximum height reached by Bullet. (3) final velocity. (4) position of bullet at t = 6 seconds. It is the problem of projectile motion.

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