Determining the time it takes to reach half of a penny's terminal speed

[aquirk]

The question is:
The terminal speed of a penny is 11 m/s. By neglecting air resistance, calculate how long it takes for a penny falling from rest to reach half of this speed

Homework Equations

I thought maybe you had to use one of the kinematic equations but that didnt work

The Attempt at a Solution

I tried Vf-Vi/a = delta T
Also, I tried dividing 11 m/s by gravity and that didnt work.
Im not sure if I am totally missing an equation needed or if I am making this way harder than it really is. Any help would be greatly appreciated!

 

[The Liberator]

you are on track with the answer, but why use 11ms-¹? (as you stated earlier that it was the terminal velocity, and you were going to use half of the terminal velocity.

Also, the equation does not need to be so complicated. Look at these ones, and figure out which one is needed, by knowing the data you have and the what is in the equations.

 

[nlightner]

I can provide a response to this question. To determine the time it takes for a penny falling from rest to reach half of its terminal speed, we need to use the equation for acceleration due to gravity, which is a = g = 9.8 m/s^2. This equation is applicable as we are neglecting air resistance in this scenario. We also need to use the equation for velocity, which is v = u + at, where v is the final velocity, u is the initial velocity, a is the acceleration, and t is the time.

In this case, the initial velocity (u) is 0 m/s, as the penny is falling from rest. The final velocity (v) is half of the terminal speed, which is 11 m/s/2 = 5.5 m/s. The acceleration (a) is the acceleration due to gravity, which is 9.8 m/s^2. We can rearrange the equation to solve for time (t):

t = (v-u)/a

Substituting the values, we get:

t = (5.5 m/s - 0 m/s)/9.8 m/s^2 = 0.56 seconds

Therefore, it takes approximately 0.56 seconds for a penny falling from rest to reach half of its terminal speed of 11 m/s, neglecting air resistance. This calculation assumes that the penny is in a vacuum and there is no air resistance acting on it. In reality, air resistance will affect the speed at which the penny falls, so this calculation may not be entirely accurate. However, it provides an estimate and gives us an idea of the time it takes for a penny to reach half of its terminal speed.