# A missile is fired at 840 m/s at an angle x to the horizontal. Prove that its range is the largest for x = 45 degree and find it. The missile is fired at 840 m/s at an angle x to the horizontal. We can divide its velocity into a horizontal component equal to 840* cos x and a vertical component equal to 840* sin x.

The vertical component starts to decrease due to the acceleration due to gravity,...

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The missile is fired at 840 m/s at an angle x to the horizontal. We can divide its velocity into a horizontal component equal to 840* cos x and a vertical component equal to 840* sin x.

The vertical component starts to decrease due to the acceleration due to gravity, reaches 0, and then reverses direction to reach – 840 * sin x. The time taken for this is (840* sin x + 840 * sin x) / 9.8

The horizontal distance travelled during the time the missile is in the air is 840*cos x*2*840*sin x / 9.8.

2*cos x* sin x = sin 2x has a maximum value of 1 when 2x = 90 or x = 45 degrees.

The range for x = 45 degrees is 840^2/9.8 = 72000 m or 72 km.

The horizontal range is maximum for x = 45 degrees and equal to 72 km.

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