Question

You toss a rock of mass m vertically upward. Air resistance can be neglected. The rock reaches a maximum height h above your hand. What is the speed of the rock when it is at height h/4? Express your answer in terms of the variables h and g.

Answer 1

Answer:

$v=\sqrt{\frac{3gh}{2}}$

Step-by-step explanation:

If air resistance is neglected, the energy is constant. That means that in every point in the trajectory the energy is the same.

At the point of height h, all the energy is potential and velocity is zero (no kinetic energy)

$E=mgh$

At point of height h/4, some of the potential energy has transformed in kinetic energy and the velocity is no longer zero. But we know the total amount of energy is the same as the maximum height point.

With that information, we can calculate the velocity v:

$E=mg(\frac{h}{4})+m\frac{v^2}{2}=mgh\\\\m\frac{v^2}{2}=mg(h-\frac{h}{4})\\\\\frac{v^2}{2}=\frac{3gh}{4}\\\\v^2=\frac{3gh}{2}\\\\v=\sqrt{\frac{3gh}{2}}$

Answer 2

Step-by-step explanation:

Below is an attachment containing the solution.