Q. An object is projected at an angle of $45{^\circ}$ with the horizontal. The horizontal range and maximum height reached will be in the ratio
A) 1:2
B) 2:1
C) 1:4
D) 4:1
Explanation
(d) $Horizontal\ range,\ \ R = \frac{u^{2}\sin^{2}45{^\circ}}{2g} = \frac{u^{2}}{g}$
$Maximum\ height,\ \ \ H = \frac{u^{2}\sin^{2}45{^\circ}}{g} = \frac{u^{2}}{4g}$
$\therefore\ \ \frac{R}{H} = \frac{4}{1}$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.