8. A 4.70 m long pole is balanced vertically on its tip. What will be the speed (in meters/second) of the tip of the pole just before it hits the ground? (Assume the lower end of the pole does not slip.)
I did V^2=0^2+2(9.8)(4.70) and was wrong. I did remember to square root at the end. My answer was 9.60 m/s.
see other post.
To determine the speed of the tip of the pole just before it hits the ground, you can use the concept of conservation of energy. At the initial position, when the pole is balanced vertically on its tip, all of its potential energy is converted to kinetic energy just before it hits the ground.
First, let's calculate the potential energy of the pole when it is balanced vertically:
Potential energy (PE) = mgh,
where m is the mass of the pole (which we can assume to be negligible), g is the acceleration due to gravity (approximately 9.8 m/s^2), and h is the height of the pole (4.70 m).
PE = mgh
≈ 0 * 9.8 * 4.70
= 0 joules
Since the pole is balanced vertically, it starts falling from rest. Therefore, its initial kinetic energy (KE) is zero.
At the end, just before the pole hits the ground, all of its potential energy is converted to kinetic energy. Therefore, we can equate the potential energy at the initial position to the kinetic energy at the final position:
PE = KE
0 = (1/2)mv^2
v^2 = 2gh
Plugging in the values:
v^2 = 2 * 9.8 * 4.70
v^2 ≈ 92.12
Now, take the square root of both sides to find the speed at the final position:
v = √(92.12)
v ≈ 9.60 m/s
Therefore, the correct answer is 9.60 m/s, confirming the result you obtained. Remember to include the square root in your calculations next time!