A cannon ball fired from top of a building at an angle 30 degree above horizontal with speed 60 m/s travels twice the horizontal distance as a cannonball projected from the same spot with same speed at an angle 30 degree below horizontal.

Let the time taken by the first cannonball to strike the ground be t. If the height of the building is H,

H = 60*sin 30*t - (1/2)*9.8*t^2

The horizontal distance traveled by the cannon ball is D = 60*cos 30*t.

Let the time taken by the second cannon ball to strike the ground be t'. As the direction of acceleration is downwards and the cannonball is also projected at 30 degrees below the horizontal H = 60*sin 30*t' + (1/2)*9.8*t'^2

The horizontal distance traveled is D' = 60*cos -30*t'

As D = 2D'

=> 60*cos 30*t = 2*60*cos -30*t'

=> t = 2*t'

60*sin 30*t - (1/2)*9.8*t^2 = 60*sin 30*t' + (1/2)*9.8*t'^2

=> 60*sin 30*2*t' - (1/2)*9.8*4*t'^2 = 60*sin 30*t' + (1/2)*9.8*t'^2

=> 60*sin 30*t' = (5/2)*9.8*t'^2

=> 60 = (5/2)*9.8*t'

=> t' = 60/49

H = 60*sin 30*t' + (1/2)*9.8*t'^2

H = 2160/49 m

**The height of the building is 2160/49 m**

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