A particle is projected from horizontal making an angle $60{^\circ}$ with initial velocity $40ms^{- 1}$. The time taken by the particle to make angel $45{^\circ}$ from horizontal, is

Question

Dr. Abubacker · Accepted Answer

(d) At $45{^\circ},v_{x} = v_{y}$
or $u_{x} = u_{y} - gt$
$	herefore\ \ t = \frac{u_{y} - u_{x}}{g}$
$= \frac{40(\sin{60{^\circ}} - \sin{30{^\circ}})}{9.8} = 1.5\ s$

Q. A particle is projected from horizontal making an angle $60{^\circ}$ with initial velocity $40ms^{- 1}$. The time taken by the particle to make angel $45{^\circ}$ from horizontal, is

Explanation

Reviewed by Dr. Abubacker

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