Question

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

Answer 1

Answer:

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

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:

$Ek=\frac{1}{2}*m*v_{0} ^{2} \\where\\v_{0} = initial velocity[m/s]$

$Ek=Ep\\\frac{1}{2} *m*v_{0} ^{2} =m*9.81*h\\h=\frac{40^{2}}{2*9.81} \\h=81.5[m]$

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.