Question

A 2 kg ball is moving 3 m/s when it starts rolling up a hill.

What is the maximum height that the ball will reach?

(ignore friction and do not include units)

Answer 1

the height reached is = 0.458 [m]

Step-by-step explanation:

We need to make a sketch of the ball and see the location of the reference point where the potential energy is zero. But the kinetic energy will be defined by the following expression:

`Ek=(1)/(2) *m*v^(2) \\where:Ek= kinetic energy [J]\\m = mass of the ball [kg]\\v = velocity of the ball [m/s]`

Replacing the values on the equation we have:

`Ek=(1)/(2)*(2)*(3^(2) )\\ Ek=9[J]\\`

This kinetic energy will be transformed in potential energy in the moment when the ball starts to rolling up. Therefore the maximum height reached by the ball depends of the initial velocity given to the ball.

`Ek=Ep\\where\\Ep=potential energy [J]\\Ep=m*g*h\\where\\g=gravity = 9.81[m/s^2]\\h=height reached [m]\\`

Now we have:

`h=(Ep)/(m*g) \\h=(9)/(2*9.81) \\\\h=0.45 [m]`

In that moment when the ball reach the 0.45 [m] the potencial energy will be maximum and equal to the kinetic energy when the ball has a velocity of 3[m/s]