Answer:
v=59[m/s]
Explanation:
To solve this problem we must use the principle of conservation of energy, which tells us that energy is transformed from Kinetic to potential or vice versa. At the moment when the car is at the top before falling down the cliff, we have the car moving at speed 50 [m/s] (kinetic energy) also it is 50 [m] above ground level (potential energy).
[tex]E_{k1}+E_{p1}=E_{k2}\\[/tex]
where:
Ek1 = kinetic energy before falling [J]
Ep1 = potential energy before falling [J]
Ek2 = kinetic energy in the ground [J]
The potential energy can be calculated by means of the following equation.
[tex]E_{p}=m*g*h[/tex]
where:
m = mass = 500 [kg]
g = gravity acceleration = 9.81 [m/s²]
h = elevation = 50 [m]
Whereas the kinetic energy can be calculated by means of the following equation.
[tex]E_{k}=\frac{1}{2}*m*v^{2}[/tex]
where:
v = velocity = 50 [m/s]
Now replacing in the general equation:
[tex]\frac{1}{2} *500*(50)^{2} +500*9.81*50=\frac{1}{2} *500*v^{2}\\625000+245250=250*v^{2} \\250*v^{2} =870250\\v=\sqrt{870250/250} \\v=59[m/s][/tex]
