Time taken to get down the wedge

$sin\theta = \frac{h}{R}$

$\boxed{h = R sin\theta}$

Using conservation of energy

$mg(20) = mg(20 - h) + \frac12mv^2$

$\boxed{v = \sqrt{2gh} = \sqrt{2gRsin\theta}}$

Now,

$v = \frac{ds}{dt}$

$v = \frac{Rd\theta}{dt}$

$\sqrt{2gRsin\theta} = \frac{Rd\theta}{dt}$

$\int_0^{t_1} dt = \sqrt{\frac{R}{2g}} \int_0^{π/2} \frac{d\theta}{\sqrt{sin\theta}}$

$t_1 = \sqrt{\frac{20}{2(10)}} × 2.62$

$\color{#3D99F6}{\boxed{t_1 = 2.62s}}$

Velocity of ball after reaching ground

$u = \sqrt{2(10)(20)sin(90°)}$

$\color{#20A900}{\boxed{u = 20m/s}}$

Time taken to cross road

$t_2 = \frac{40}{u}$

$\color{#3D99F6}{\boxed{t_2 = 2s}}$

Time taken to climb second wedge

By symmetry

$t_3 = t_1$

$\color{#3D99F6}{\boxed{t_3 = 2.62s}}$

Time period

$T = 2(t_1 + t_2 + t_3)$

$\color{#3D99F6}{\boxed{T = 14.48s}}$