Answer: 0.625 m
Explanation:
Given:
Final velocity of the car = 3.5 m/s
While the car rolls down the hill, there is a conversion of potential energy (PE) to kinetic energy (KE). Therefore, applying the conversion of energy as,
[tex]$Total energy at height $(\mathrm{h})=$ Total energy at bottom$\begin{aligned}&K E+P E=K E^{\prime}+P E^{\prime} \\&0+m g h=\frac{1}{2} m v^{2}+O \\&m g h=\frac{1}{2} m v^{2}\end{aligned}$[/tex]
Here, m denotes the mass of the car, g is the gravitational acceleration, having a value of 9.8 m/s^2. And h is the height of the hill.
Solving for h,
[tex]\begin{aligned}&g h=\frac{l}{2} v^{2} \\&h=\frac{1}{2} \times \frac{v^{2}}{g} \\&h=\frac{1}{2} \times \frac{3.5^{2}}{9.8} \\&h=0.625 \mathrm{~m}\end{aligned}[/tex]
Therefore, the required height of hill is 0.625 m
