\(t_1=\frac{\sqrt2⋅\frac{H}{2}}{g}=\frac{\sqrt{H}}{g}\)
\(t_2=\frac{\sqrt{2H}}{g}−t_1\)
\(⇒t_2=\frac{\sqrt{2H}}{g}−\frac{\sqrt{H}}{g}\)
\(⇒t2=√Hg{√2−1}\)
\(⇒t2=(√2−1)t1\)
So, the correct option is (D): t2 =(√2−1)t1
Let \( y = f(x) \) be the solution of the differential equation
\[ \frac{dy}{dx} + 3y \tan^2 x + 3y = \sec^2 x \]
such that \( f(0) = \frac{e^3}{3} + 1 \), then \( f\left( \frac{\pi}{4} \right) \) is equal to:
Find the IUPAC name of the compound.
If \( \lim_{x \to 0} \left( \frac{\tan x}{x} \right)^{\frac{1}{x^2}} = p \), then \( 96 \ln p \) is: 32
Formula to find distance between two parallel line:
Consider two parallel lines are shown in the following form :
\(y = mx + c_1\) …(i)
\(y = mx + c_2\) ….(ii)
Here, m = slope of line
Then, the formula for shortest distance can be written as given below:
\(d= \frac{|c_2-c_1|}{\sqrt{1+m^2}}\)
If the equations of two parallel lines are demonstrated in the following way :
\(ax + by + d_1 = 0\)
\(ax + by + d_2 = 0\)
then there is a little change in the formula.
\(d= \frac{|d_2-d_1|}{\sqrt{a^2+b^2}}\)