Question

An object is projected with a speed u. If v is the speed at the highest position, then angle of
projection is ____​

Answer 1

Answer:

B is correct .............

Answer 2

Answer:

Let the angle of projection be alpha , initial speed is u . X coordinate of u is

$= u \cos( \alpha )$

At highest point only x coordinate remains and y coordinate will be zero , so

$v = u \cos( \alpha ) \\ \\ \alpha = \cos {}^{ - 1} ( \frac{v}{u} ) = \sin { }^{ - 1} ( \frac{ \sqrt{ {u}^{2} - {v}^{2} } } {u}) \\ \\ = \tan {}^{ - 1} ( \frac{ \sqrt{ {v}^{2} - {u}^{2} } }{v} )$

Option 4th