Answer
The maximum angle for the hill is $9.9^{\circ}$
Work Step by Step
We can use kinematics to find the acceleration of the car.
$a = \frac{\Delta v}{t} = \frac{21 ~m/s}{12.5 ~s} = 1.68 ~m/s^2$
We can use a force equation to find the force $F_p$ which can be exerted to push the car forward:
$\sum F = ma$
$F_p = (920 ~kg)(1.68 ~m/s^2) = 1546 ~N$
If the car does not slow down while climbing the hill, then let's suppose that the acceleration is zero. Then we can assume that $F_p$ is equal in magnitude to the component of the car's weight which is directed down the hill as the car climbs up the hill:
$mg ~sin(\theta) = F_p$
$sin(\theta) = \frac{F_p}{mg} = \frac{1546 ~N}{(920 ~kg)(9.80 ~m/s^2)}$
$\theta = sin^{-1}(\frac{1546 ~N}{(920 ~kg)(9.80 ~m/s^2)}) = 9.9^{\circ}$
The maximum angle for the hill is $9.9^{\circ}$
