JEE-Mains 2014 (29/30)

Geometry Level 2

A bird is sitting on the top of a vertical pole 20 m 20 \text{ m} high and its elevation from a point O O on the ground is 4 5 ∘ 45^{\circ} . If flies off horizontally straight away from the point O O . After one second, the elevation of the bird from O O is reduced to 3 0 ∘ 30^{\circ} . Then the speed (in m/s) of the bird is :

40 ( 2 − 1 ) 40(\sqrt{2}-1) 20 2 20\sqrt{2} 20 ( 3 − 1 ) 20(\sqrt{3}-1) 40 ( 3 − 2 ) 40(\sqrt{3}-\sqrt{2})

View wiki
Sandeep Bhardwaj by Sandeep Bhardwaj , 28, India

This section requires Javascript.
You are seeing this because something didn't load right. We suggest you, (a) try refreshing the page, (b) enabling javascript if it is disabled on your browser and, finally, (c) loading the non-javascript version of this page . We're sorry about the hassle.

0 solutions

No explanations have been posted yet. Check back later!

0 pending reports

×

Problem Loading...

Note Loading...

Set Loading...