rob colley set a record in “pole sitting” when he spent 42 days in a barrel at the top of a flagpole with a height of 43 meters, suppose you wanted to throw him up a sandwich. with what speed do you have to throw the sandwich up so that it would just reach the barrel?
v^2 = 2as = 19.6 * 43 = 842.8
v = 29.03 m/s
check:
h = 29.03t - 4.9t^2
the vertex is at (2.96,43)
To calculate the speed at which you need to throw the sandwich up so that it just reaches the barrel, you can use the principle of conservation of mechanical energy. This principle states that the sum of the potential energy and kinetic energy of an object remains constant, assuming there are no other external forces acting on it.
First, let's determine the potential energy of the sandwich when it reaches the barrel at the top of the flagpole. The potential energy (PE) is given by the formula:
PE = m * g * h
where m is the mass of the sandwich, g is the acceleration due to gravity (approximately 9.8 m/s^2), and h is the height of the flagpole (43 meters).
Next, let's determine the kinetic energy of the sandwich right after you throw it upward. The kinetic energy (KE) is given by the formula:
KE = (1/2) * m * v^2
where v is the velocity (speed) with which you throw the sandwich upward.
Since the sandwich starts from rest at the bottom of the flagpole, its initial kinetic energy is zero. Therefore, the sum of the potential energy and kinetic energy at the top should equal the potential energy when it reaches the barrel.
PE + KE = PE
m * g * h + 0 = m * g * h
To find the velocity (v), we can rearrange this equation:
0 = m * g * h - m * g * h
0 = m * g * h * (1 - 1)
0 = 0
Since the equation doesn't provide any meaningful information, it means that throwing the sandwich with any speed would allow it to reach the barrel as long as it is thrown vertically upward.
However, in reality, air resistance and other factors may affect the motion of the sandwich, so it will be challenging to precisely determine the exact speed required without considering these additional factors.
To solve this problem, we can use the principle of conservation of energy, assuming that there is no air resistance.
1. First, let's find the potential energy of the sandwich at the top of the flagpole. The potential energy is given by the equation:
Potential Energy (PE) = mass (m) * acceleration due to gravity (g) * height (h)
Given that the height of the flagpole is 43 meters, we can assume the sandwich is at ground level initially, so the height (h) is 43 meters. The mass (m) of the sandwich is not provided, so we can assume it to be 0.1 kg and acceleration due to gravity (g) is approximately 9.8 m/s^2.
Potential Energy (PE) = 0.1 kg * 9.8 m/s^2 * 43 m
= 42.14 Joules
2. Next, let's find the kinetic energy that the sandwich needs at the bottom of the throw (ground level). The kinetic energy is given by the equation:
Kinetic Energy (KE) = 0.5 * mass (m) * velocity (v)^2
We want the sandwich to just reach the top of the flagpole, so at that point, its kinetic energy should be equal to the potential energy calculated above.
0.5 * 0.1 kg * v^2 = 42.14 Joules
3. Rearranging the equation, we can solve for the velocity (v):
v^2 = (2 * PE) / m
= (2 * 42.14 Joules) / 0.1 kg
= 842.8 m^2/s^2
v ≈ √842.8 m^2/s^2
≈ 29.02 m/s
Therefore, you would need to throw the sandwich up with a speed of approximately 29.02 m/s for it to just reach the barrel at the top of the flagpole.