Answer:
Final speed, v = 54.64 m/s.
Explanation:
Given that,
Mass of a car, m = 2000 kg
Initial speed of the car, u = 23 mph = 10.28 m/s
height, h = 144 m
We need to find the speed of the car at the bottom of the hill. Let v is the speed. Applying conservation of energy as :
[tex]\dfrac{1}{2}mu^2+mgh=\dfrac{1}{2}mv^2\\\\v^2-u^2=2ah[/tex]
a = g here
[tex]v=\sqrt{2gh+u^2}\\\\v=\sqrt{2\times 10\times 144+(10.28)^2}\\\\v=54.64\ m/s[/tex]
So, the speed at the bottom of the hill is 54.64 m/s.