Sunday, October 27, 2013

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, 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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