Question

Mr. Colley set a record in ""pole sitting"" where he spent 42 days in a barrel at the top of a flagpole that has a height of 43 meters. Suppose a friend who wants to deliver an ice-cream sandwich to Mr. Colley throws the ice cream sandwich up to him with just enough speed to reach the barrel. How long does it take the ice-cream sandwich to reach the barrel?

Answer 1

The ice-cream sandwich takes approximately 2.1 seconds to reach Mr. Colley in the barrel at the top of the flagpole.

Step-by-step explanation:

To calculate the time it takes for the ice-cream sandwich to reach the barrel, we can use the principles of projectile motion. The projectile will follow a parabolic trajectory. Since the initial vertical velocity is zero and the only force acting on it is gravity, we can use the equation h = (1/2)gt^2 to find the time it takes for the ice-cream sandwich to reach the barrel.

First, we need to find the value of g, the acceleration due to gravity. On Earth, g is approximately 9.8 m/s^2. The height of the flagpole is 43 meters. Plugging these values into the equation, we get 43 = (1/2)(9.8)t^2. Solving for t, we find that it takes the ice-cream sandwich approximately 2.1 seconds to reach the barrel.

Therefore, the ice-cream sandwich takes approximately 2.1 seconds to reach Mr. Colley in the barrel at the top of the flagpole.

Answer 2

To calculate the time it takes for an ice cream sandwich to reach the top of a 43-meter flagpole, we use kinematics equations for uniformly accelerated motion, considering the acceleration due to gravity. By setting the displacement to 43 meters and rearranging the kinematic equation, the time can be estimated at approximately 2.96 seconds, ignoring air resistance and other possible factors.

Step-by-step explanation:

Calculating Time for an Ice-Cream Sandwich to Reach the Top of a Flagpole

To solve how long it takes for an ice cream sandwich to reach a barrel at the top of a 43-meter flagpole, we need to apply the principles of kinematics, specifically the equations of motion for uniformly accelerated linear motion. The acceleration is due to gravity and is directed downwards. We assume the initial vertical velocity is just enough to reach 43 meters, so there is no leftover velocity at that height. This means that at the peak of its trajectory, the velocity of the ice-cream sandwich is zero.

One of the kinematic equations that relates displacement (s), initial velocity (u), acceleration (a), and time (t) is:

s = ut + (1/2)at2

In this case, the displacement (s) is 43 meters, initial velocity (u) is unknown, acceleration (a) is -9.8 m/s2 (negative because it is downward), and time (t) is what we are solving for. Rearranging the equation to solve for the time gives:

t = (sqrt(2*s/a)

Plug in the values:

t = (sqrt(2*43 m/9.8 m/s2))

Calculating this gives a time of about 2.96 seconds for the ice-cream sandwich to reach Mr. Colley. It is crucial to remember that this is an idealized scenario that ignores air resistance and assumes that the sandwich is projected vertically with sufficient initial speed to just reach 43 meters without any horizontal component of motion.