Question bank/JEE Main/Physics

A particle at a height from the ground is projected with
an angle 30º fromthe horizontal, it strikes the ground making angle 45º
with horizontal. It is again projected from the same point with the same
speed but with an angle of 60º with horizontal. The tangent of the angle
it makes with the horizontal is $\sqrt{n}$when it strikes the ground

Your answerCheck answer

Show answer and solution▾

Using $v = \sqrt{u^{2} + 2gh}$
$v = \sqrt{u^{2}\sin^{2}\theta + 2gh}$(Vertical comp. when striking)
$ucos\theta = \sqrt{u^{2}\sin^{2}\theta + 2gh}$
$u^{2}\cos^{2}\theta = u^{2}\sin^{2}\theta + 2gh$
$u^{2}\left( \frac{3}{4} - \frac{1}{4} \right) = 2gh$
$u^{2} = 4gh$
$u = 2\sqrt{gh}$

Source: Grass-G / P-05 Physics paper 10 October.docx