Question

A roller coaster starts from rest at its highest point and then descends on its (frictionless) track. its speed is 30 m/s when it reaches ground level.

Answer 1

Missing question in the text. Found on internet:
"What was its speed when its height was half that of its starting point?"

Solution:
we can solve the problem by using the law of conservation of energy.

At ground level (let's label this point as point A), the roller coaster has only kinetic energy (because its height is zero, so its potential energy is zero as well), which is
$E_A=K_A = \frac{1}{2} mv_A^2$
where m is the mass of the roller coaster and vA is its speed at ground level.

When the roller coaster is at half of the starting height, it has both kinetic energy and potential energy:
$E_B = K_B+U_B = \frac{1}{2}mv_B^2 + mg \frac{h}{2}$
where h is the initial height at the starting point.

Since the energy must be conserved, we can write
$E_A = E_B$
therefore
$\frac{1}{2}mv_A^2 = \frac{1}{2} mv_B^2 + \frac{1}{2} mgh$
and so
$v_B^2 = v_A^2 - gh$ (1)

We know vA=30 m/s, however, we don't know the height h at the starting point. But we can calculate it by applying again the conservation of energy. At its starting point (let's label it point C), the roller coaster has only potential energy, because it starts from rest:
$E_C = U_C = mgh$
The total energy must be equal to the total energy at ground level (point A), so
$E_A = E_C$
$\frac{1}{2}mv_A^2 = mgh$
from which we find
$h= \frac{v_A^2}{2g}= \frac{(30 m/s)^2}{2 \cdot 9.81 m/s^2} =45.9 m$

And now we can substitute h into (1), to find the velocity of the roller coaster in point B:
$v_B = \sqrt{v_A^2 - gh} = \sqrt{(30 m/s)^2-(9.81 m/s^2)(45.9m)} =21.2 m/s$