We can solve this problem using conservation of energy. At the bottom of the dip, all of the car's initial energy is in the form of kinetic energy, since it is moving and has no height. At the point where the speed is zero, all of the car's energy is in the form of potential energy, since it has no speed but has gained height.
We can use the conservation of energy equation:
$E_i = E_f$
where $E_i$ is the initial energy and $E_f$ is the final energy. In this case, we can write:
$\frac{1}{2}mv_i^2 + mgh_i = \frac{1}{2}mv_f^2 + mgh_f$
where $m$ is the mass of the car, $v_i$ is the initial speed (at the bottom of the dip), $h_i$ is the initial height (which is zero), $v_f$ is the final speed (when the car has reached its highest point and its speed is zero), and $h_f$ is the final height (which is what we are trying to find).
We can plug in the given values:
$\frac{1}{2}(500\text{ kg})(16\text{ m/s})^2 + (500\text{ kg})(9.8\text{ m/s}^2)(0\text{ m}) = \frac{1}{2}(500\text{ kg})(0\text{ m/s})^2 + (500\text{ kg})(9.8\text{ m/s}^2)h_f$
Simplifying, we get:
$64000\text{ J} = 4900\text{ J}h_f$
Solving for $h_f$, we get:
$h_f = \frac{64000\text{ J}}{4900\text{ J}} \approx 13.06\text{ m}$
Therefore, the magnitude of the vertical height above the lowest point, when the speed is zero, is approximately 13.06 m.
A roller-coaster car has a mass of 500 kg when fully loaded with passengers. At the bottom of a circular dip of radius 40 m (as shown in the figure) the car has a speed of 16 m/s. What is the magnitude of the vertical height above the lowest point, when the speed is zero? Ignore friction.
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