A very silly person, intent on catching pigeons on the roof of an apartment building, trips and falls a distance of 43.0 m. She lands on a metal garbage can, crushing it to a depth of 0.457 m and walks away without having been seriously hurt. What acceleration did she experience during the collision?

Assuming the top of the garbage can is 43 m below the roof.

Acceleration due to gravity, a = -g = -9.8 m/s/s
Distance, S = -43 m
initial velocity, u = 0 m/s
final velocity, v
v²-u²=2aS
v=sqrt(2aS+u²)
=sqrt(2*(-9.8)*(-43))
=29.03 m/s
For impact with garbage can,
u=29.03 m/s
v=0
S=-0.457 m
Using the same formula
v²-u²=2aS
Can you find the acceleration a?

The acceleration is 922 m/s^2.

How come v=0 in the final equation? I thought that v represented the velocity at impact rather than the velocity after the girl lands.

I was using the same terminology, namely

u=initial, v=final,
that is why v=0 for the decceleration part.
u=initial equals the velocity attained during the free-fall, namely 29.03 m/s as indicated.
922 m/s/s is correct, considering that it took 43 m to accelerate to 29 m/s, and is reduced to zero in 0.457 m.

I don't understand how you got that the acceleration is 922 m/s/s. I've been trying to understand it for a long time and it's not making sense.

I will rephrase the last part:

For impact with garbage can,
Velocity on impact with garbage can, u = 29.03 m/s
Final velocity (at rest), v = 0
Displacement, equal to denting of garbage can, S = -0.457 m
Using the same formula
v²-u²=2aS
a=(v²-u²)/2S
=(0²-29.03²)/(2*0.457)
=842.74/(.914)
=922.0 m/s/s

Hello.

I was just wondering...
Since the question said "She lands on a metal garbage can, crushing it to a depth of 0.457 m", how did you know that final depth of the garbage can was equal to the negative displacement caused by crushing?

To find the acceleration experienced during the collision, we can use the equation of motion:

\[ v^2 = u^2 + 2a \cdot s \]

Where:
- \( v \) is the final velocity of the person.
- \( u \) is the initial velocity of the person (which we assume to be zero).
- \( a \) is the acceleration experienced during the collision.
- \( s \) is the distance traveled during the collision.

In this scenario, the person falls from a distance of 43.0 m and stops after crushing the garbage can to a depth of 0.457 m. Therefore, the distance traveled during the collision (\( s \)) is the sum of the distance fallen and the depth crushed:

\[ s = 43.0 \, \text{m} + 0.457 \, \text{m} = 43.457 \, \text{m} \]

Since the initial velocity (\( u \)) is zero, we can simplify the equation to:

\[ v^2 = 2a \cdot s \]

Now, we need to find the velocity (\( v \)). To do this, we can use the concept of conservation of energy, assuming no energy loss during the fall. The potential energy gained when the person falls is converted into kinetic energy just before impact:

\[ m \cdot g \cdot h = \frac{1}{2} \cdot m \cdot v^2 \]

Where:
- \( m \) is the mass of the person. (Note: The mass was not provided, so we can't find the exact acceleration, but we can still find the expression for acceleration.)
- \( g \) is the acceleration due to gravity (approximately 9.8 m/s^2).
- \( h \) is the height from which the person falls.

Knowing that the height fallen (\( h \)) is 43.0 m, we can manipulate the equation and substitute the value of \( v^2 \):

\[ 43.0 \, \text{m} \cdot 9.8 \, \text{m/s}^2 = \frac{1}{2} \cdot m \cdot v^2 \]

Solving for \( v^2 \), we get:

\[ v^2 = \frac{43.0 \cdot 9.8}{0.5} \, \text{m}^2/\text{s}^2 \]

Now we have both \( v^2 \) and \( s \), so we can substitute these values back into \( v^2 = 2a \cdot s \) to find the acceleration (\( a \)):

\[ \frac{43.0 \cdot 9.8}{0.5} = 2a \cdot 43.457 \]

Solving for \( a \), we get:

\[ a = \frac{43.0 \cdot 9.8}{0.5 \cdot 43.457} \, \text{m/s}^2 \]

Calculating this expression will give you the acceleration experienced during the collision.