The coefficient of friction between a road and the tyre of a...

Initial speed of the vehicle, $u = 36 \text{ km/hr} = 36 \times \dfrac{5}{18} = 10 \text{ m/s}$

Final speed, $v = 0$

Stopping distance, $s = 5 \text{ m}$

Using the equation of motion $v^2 = u^2 - 2as$, we can find the required deceleration $a$:

$0 = (10)^2 - 2a(5)$

$10a = 100 \Rightarrow a = 10 \text{ m/s}^2$

When the vehicle is moving downwards on the incline, the component of gravity acts downwards and the kinetic friction acts upwards. The net retardation is given by:

$a = \dfrac{\mu mg \cos \theta - mg \sin \theta}{m} = g(\mu \cos \theta - \sin \theta)$

Substituting $a = 10 \text{ m/s}^2$, $g = 10 \text{ m/s}^2$, and $\mu = \dfrac{4}{3}$:

10 = 10 \left(\dfrac{4}{3} \cos \theta - \sin \theta \right)

$1 =$\dfrac{4}{3} \cos \theta - \sin \theta

$3 = 4 \cos \theta - 3 \sin \theta$

Dividing both sides by $5$:

\dfrac{4}{5} \cos \theta - \dfrac{3}{5} \sin \theta = \dfrac{3}{5}

$\cos 37^\circ \cos \theta - \sin 37^\circ \sin \theta = \cos 53^\circ$

$\cos(\theta + 37^\circ) = \cos 53^\circ$

$\theta + 37^\circ = 53^\circ$

$\theta = 16^\circ$