A 2.5m long pole is balanced vertically on it's tip. If it falls over with it's bottom never slipping. Find the speed of the top end of the pole just before it hits the ground.
To find the speed of the top end of the pole just before it hits the ground, we can consider the conservation of energy. At the topmost position, all of the potential energy is in the pole, and as it falls, this potential energy is converted into kinetic energy.
Let's denote the height of the top end of the pole from the ground as h. At the topmost position, the potential energy is given by:
Potential energy = m * g * h
where m is the mass of the pole, and g is the acceleration due to gravity.
When the top end of the pole hits the ground, all of the potential energy is converted into kinetic energy. The kinetic energy is given by:
Kinetic energy = (1/2) * m * v^2
where v is the speed of the top end of the pole just before it hits the ground.
Since we have the conservation of energy, we can equate the potential energy and kinetic energy equations:
m * g * h = (1/2) * m * v^2
Simplifying and solving for v, we get:
v^2 = 2 * g * h
v = sqrt(2 * g * h)
Now, we need to determine the value of h, the height of the top end of the pole at the moment just before it hits the ground.
Since the pole is balanced vertically on its tip, it forms a right-angled triangle with the ground. The length of the pole, 2.5m, is the hypotenuse of this triangle. The height, h, can be found using the Pythagorean theorem:
h^2 + (0.5)^2 = (2.5)^2
h^2 + 0.25 = 6.25
h^2 = 6
h = sqrt(6)
Substituting the value of h into the equation for v, we get:
v = sqrt(2 * g * sqrt(6))
Now, to determine the value of g, we need to know the acceleration due to gravity. The average value for g on Earth is approximately 9.8 m/s^2.
Finally, substituting the value of g and evaluating the expression, we can find the speed of the top end of the pole just before it hits the ground.
V^2 = Vo^2 + 2g*h.
V^2 = 0 + 19.6*2.5 = 49
V = 7 m/s.