- Home
- Standard 11
- Physics
3-2.Motion in Plane
normal
Two bodies are projected with the same velocity. If one is projected at an angle of ${30^o}$ and the other at an angle of ${60^o}$ to the horizontal, the ratio of the maximum heights reached is
A$3:1$
B$1:3$
C$1:2$
D$2:1$
Solution
(b) As $H = \frac{{{u^2}{{\sin }^2}\theta }}{{2g}}$
$⇒$ $\frac{{{H_1}}}{{{H_2}}} = \frac{{{{\sin }^2}{\theta _1}}}{{\sin {\theta _2}}} = \frac{{{{\sin }^2}30^\circ }}{{{{\sin }^2}60}}$= $\frac{{1/4}}{{3/4}} = \frac{1}{3}$
$⇒$ $\frac{{{H_1}}}{{{H_2}}} = \frac{{{{\sin }^2}{\theta _1}}}{{\sin {\theta _2}}} = \frac{{{{\sin }^2}30^\circ }}{{{{\sin }^2}60}}$= $\frac{{1/4}}{{3/4}} = \frac{1}{3}$
Standard 11
Physics
