Calculating Clay's Acceleration: 12.0 m Fall

[lgen0290]

Homework Statement

A ball of moist clay falls 12.0 m to the ground. It is in contact with the ground for 22.0 ms before stopping.
(a) What is the average acceleration of the clay during the time it is in contact with the ground? (Treat the ball as a particle.)

Homework Equations

v^2 = u^2 + 2gh
v=final speed
u= initial speed

The Attempt at a Solution

I'mm not sure where to start this, as I don't think it is asking about the whole thing, but the 22 ms? I'm lost here.

 

[h2oski1326]

Ok, well first you could figure out how fast the ball is going when it hits the ground.

The next step is to ask yourself what the definition of average acceleration is and see if you have all the information you need to solve for it.

 

[Moetasim]

As a scientist, your first step would be to carefully read and understand the problem. In this case, the problem is asking for the average acceleration of the clay during the time it is in contact with the ground. The given information is that the clay falls 12.0 m and is in contact with the ground for 22.0 ms (milliseconds) before stopping.

To calculate acceleration, we can use the equation a = (vf - vi)/t, where a is acceleration, vf is final velocity, vi is initial velocity, and t is time. In this case, we are treating the clay as a particle, so we can ignore any effects of its shape or size.

We are given the distance the clay falls, 12.0 m, and the time it is in contact with the ground, 22.0 ms. However, we need to find the initial and final velocities in order to use the equation.

To find the initial velocity, we can use the equation v^2 = u^2 + 2gh, where v is the final velocity, u is the initial velocity, g is the acceleration due to gravity (9.8 m/s^2), and h is the height the clay falls. Plugging in the given values, we get:

v^2 = u^2 + 2(9.8 m/s^2)(12.0 m)
v^2 = u^2 + 235.2 m^2/s^2

Since the clay is initially at rest (u = 0), we can solve for v:

v^2 = 0 + 235.2 m^2/s^2
v = √235.2 m/s
v = 15.33 m/s

Now that we have the initial and final velocities, we can use the acceleration equation to find the average acceleration of the clay:

a = (vf - vi)/t
a = (15.33 m/s - 0 m/s)/22.0 ms
a = 0.697 m/s^2

Therefore, the average acceleration of the clay during the time it is in contact with the ground is 0.697 m/s^2.