Question

A car with a mass of 500 kg, gets off road and goes straight into a cliff of 30 m with an
initial velocity of 50 m/s. What is the car velocity when it hits the ground?
I

Answer 1

Answer:

The velocity of the car when it hits the ground is approximately 55.574 meters per second.

Explanation:

According to this expression, the car goes straight into the cliff and goes down due to gravity, whose situation is described by the Principle of Energy Conservation and supposing that all non-conservative forces are negligible:

$U_{g,1}+K_{1} = U_{g,2}+K_{2}$ (1)

Where:

$U_{g,1}$, $U_{g,2}$ - Gravitational potential energies of the car at the top and the bottom, measured in joules.

$K_{1}$, $K_{2}$ - Translational kinetic energies of the car at the top and the bottom, measured in joules.

By definitions of gravitational potential and translational kinetic energies, we expand and simplify the equation above:

$\frac{1}{2}\cdot m\cdot v_{2}^{2} = \frac{1}{2}\cdot m \cdot v_{1}^{2}+m\cdot g \cdot (z_{1}-z_{2})$ (2)

$v_{2}^{2} = v_{1}^{2}+2\cdot g\cdot (z_{1}-z_{2})$

$v_{2} = \sqrt{v_{1}^{2}+2\cdot g\cdot (z_{1}-z_{2})}$

Where:

$m$ - Mass, measured in kilograms.

$v_{1}$, $v_{2}$ - Velocity of the car at the top and the bottom, measured in meters per second.

$g$ - Gravitaitional acceleration, measured in meters per square second.

$z_{1}$, $z_{2}$ - Height of the car at the top and at the bottom, measured in meters.

If we know that $v_{1} = 50\,\frac{m}{s}$, $g = 9.807\,\frac{m}{s^{2}}$, $z_{1} = 30\,m$ and $z_{2} = 0\,m$, then the velocity of the car when it hits the ground is:

$v_{2} = \sqrt{\left(50\,\frac{m}{s} \right)^{2}+2\cdot \left(9.807\,\frac{m}{s^{2}} \right)\cdot (30\,m-0\,m)}$

$v_{2}\approx 55.574\,\frac{m}{s}$

The velocity of the car when it hits the ground is approximately 55.574 meters per second.