\(t_1=\frac{\sqrt2⋅\frac{H}{2}}{g}=\frac{\sqrt{H}}{g}\)
\(t_2=\frac{\sqrt{2H}}{g}−t_1\)
\(⇒t_2=\frac{\sqrt{2H}}{g}−\frac{\sqrt{H}}{g}\)
\(⇒t2=√Hg{√2−1}\)
\(⇒t2=(√2−1)t1\)
So, the correct option is (D): t2 =(√2−1)t1
Let $ f: \mathbb{R} \to \mathbb{R} $ be a twice differentiable function such that $$ f''(x)\sin\left(\frac{x}{2}\right) + f'(2x - 2y) = (\cos x)\sin(y + 2x) + f(2x - 2y) $$ for all $ x, y \in \mathbb{R} $. If $ f(0) = 1 $, then the value of $ 24f^{(4)}\left(\frac{5\pi}{3}\right) $ is:
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.
\(d= \frac{|d_2-d_1|}{\sqrt{a^2+b^2}}\)