Question

A baseball team is practicing throwing balls vertically upward to test their throwing arms. A player manages to throw a ball that reaches a maximum altitude of 10 m above the launch point, before falling back down. The acceleration due to gravity is 9.80 m/s 2 . (a) With what initial speed was the ball thrown?

Answer 1

We can solve this problem using the law of conservation of energy.
This law states that energy in a closed system must stay same.
That means that the energy of a ball leaving the hand and the energy of a ball when it reaches its maximum height must be the same.
The energy of a ball leaving the players hand is kinetic energy:
$E_k=\frac{mv^2}{2}$
The energy when the ball reaches its maximum height ( and has zero velocity) is potential energy in a gravitational field:
$E_p=mgh$
As said before these energies must be the same, and that allows us to find the initial speed:
$E_k=E_p\\ \frac{mv^2}{2}=mgh\\ v^2=2gh\\ v=\sqrt{2gh}$
When we plug in all the number we get that $v=14\frac{m}{s}$