Answer:
6.82842712475
Explanation:
t = Time taken
u = Initial velocity
v = Final velocity
s = Displacement
a = Acceleration
g = Acceleration due to gravity = 9.81 m/s²
m denotes max
b denotes below
a denotes above
[tex]v^2-u^2=2as\\\Rightarrow -u^2=2as-v^2\\\Rightarrow u=\sqrt{v^2-2as}\\\Rightarrow u=\sqrt{0^2-2(-g)y_m}\\\Rightarrow u=\sqrt{2gy_m}\ m/s[/tex]
[tex]s=ut+\frac{1}{2}at^2\\\Rightarrow 0=ut_a+\frac{1}{2}\times g\times t_a^2\\\Rightarrow t_a=\dfrac{2u}{g}[/tex]
Substituting u in the above equation
[tex]t_a=\dfrac{2\sqrt{2gy_m}}{g}[/tex]
Now
[tex]\dfrac{y_m}{2}=ut_b-\dfrac{1}{2}gt_b^2\\\Rightarrow gt_b^2-2ut_b+y_m=0[/tex]
Solving the equation
[tex]t_b=\dfrac{2u\pm\sqrt{4u^2-4gy_m}}{2g}\\\Rightarrow t_b=\dfrac{u\pm\sqrt{u^2-gy_m}}{g}\\\Rightarrow t_b=\dfrac{\sqrt{2gy_m}\pm\sqrt{u^2-gy_m}}{g}\\\Rightarrow t_b=\dfrac{\sqrt{2gy_m}+\sqrt{u^2-gy_m}}{g}\\\Rightarrow t_b=\dfrac{\sqrt{gy_m}}{g}(\sqrt{2}-1)[/tex]
The ratio is
[tex]\dfrac{t_a}{t_b}=\dfrac{\dfrac{2\sqrt{2gy_m}}{g}}{\dfrac{\sqrt{gy_m}}{g}(\sqrt{2}-1)}\\\Rightarrow \dfrac{t_a}{t_b}=\dfrac{2\sqrt{2}}{\sqrt{2}-1}\\\Rightarrow \dfrac{t_a}{t_b}=6.82842712475[/tex]
The ratio is 6.82842712475
