Question

3. A ball is thrown straight up with an initial velocity of 40 m/s.

A. What is its velocity at the top of its path?

Answer 1

The velocity at the top of its path will be zero (0)

Step-by-step explanation:

We can solve this problem or particular situation using the principle of energy conservation.

Which tells us that energy is transformed from kinetic energy to potential energy and vice versa. A reference point should be considered at which the potential energy is zero, and at this point the initial velocity of 40 [m/s] is printed to the ball.

`Ek=Ep\\where:\\Ek=kinetic energy [J]\\Ep=potencial energy [J]`

The potential energy is determined by:

`Ep=m*g*h\\where:\\m=mass of the ball[kg}\\g=gravity[m/s^2]\\h=heigth [m]\\`

The kinetic energy is determined by:

`Ek=(1)/(2)*m*v_(0) ^(2)  \\where\\v_(0) = initial velocity[m/s]`

`Ek=Ep\\(1)/(2) *m*v_(0) ^(2) =m*9.81*h\\h=(40^(2))/(2*9.81) \\h=81.5[m]`

This will be the maximum path but, its velocity at this point will be zero. Because now all the kinetic energy has been transformed in potential energy.