# A cage descends a mine shaft with an acceleration of 0.5 m/s2. After the cage has travelled 25 metres,

a stone is dropped from the top of the shaft. Determine : 1. the time taken by the stone to hit the cage,
and 2. distance travelled by the cage before impact.

## 1. Well, the time taken by the stone to hit the cage can be determined by the equation of motion for the stone: s = ut + (1/2)at^2

Let's assume the initial velocity of the stone is zero (u = 0), and the acceleration (a) is equal to the acceleration of the cage, which is 0.5 m/s^2. Also, the distance traveled by the stone (s) is 25 meters, so we can write the equation as:
25 = 0 + (1/2)(0.5)t^2

Solving for t, we get:
50 = 0.5t^2
t^2 = 100
t = √100
t = 10 seconds

So, it will take the stone 10 seconds to hit the cage.

2. Now, to determine the distance traveled by the cage before impact, we can use the equation of motion for the cage: s = ut + (1/2)at^2
The initial velocity (u) is zero, and the acceleration (a) is 0.5 m/s^2. We need to find the distance traveled by the cage (s).

Let's substitute the values into the equation:
s = 0 + (1/2)(0.5)(10)^2
s = 0.25 * 100
s = 25 meters

So, the cage will travel a distance of 25 meters before the stone hits it.

Didn't "shaft" you with my answer, did I?

## To solve this problem, we can use the equations of motion to find the time taken by the stone to hit the cage and the distance travelled by the cage before impact. Let's solve each part step by step:

1. Time taken by the stone to hit the cage:
We know that the acceleration due to gravity (g) is approximately 9.8 m/s^2, and the initial velocity (u) of the stone when it is dropped is 0 m/s.

We can use the following equation of motion:
s = ut + (1/2)at^2

Where:
s = distance
u = initial velocity
a = acceleration
t = time

Substituting the values into the equation, we get:
25 = 0 + (1/2) * 9.8 * t^2

Simplifying the equation:
25 = 4.9t^2

Dividing both sides of the equation by 4.9:
t^2 = 25 / 4.9

Taking the square root of both sides:
t = √(25 / 4.9)
t ≈ 3.187 seconds (rounded to three decimal places)

Therefore, the time taken by the stone to hit the cage is approximately 3.187 seconds.

2. Distance travelled by the cage before impact:
To find the distance travelled by the cage (d) before impact, we can use the equation of motion:
d = ut + (1/2)at^2

In this case, the initial velocity of the cage is unknown. However, we know that the cage starts with an acceleration of 0.5 m/s^2.

Since the initial velocity is not given, we can assume it to be zero (u = 0) as the cage is descending.

Therefore, the equation simplifies to:
d = (1/2) * a * t^2

Substituting the known values:
d = (1/2) * 0.5 * (3.187)^2

Calculating the equation:
d ≈ 1.278 meters (rounded to three decimal places)

Therefore, the distance travelled by the cage before impact is approximately 1.278 meters.

## To determine the time taken by the stone to hit the cage and the distance traveled by the cage before impact, we need to analyze the motion of both the cage and the stone separately.

1. Time taken by the stone to hit the cage:
We can use the kinematic equation for free fall motion to calculate this. The equation is:

s = ut + (1/2)at^2

where:
s is the displacement (25m for the stone)
u is the initial velocity (0m/s for the stone as it is dropped)
a is the acceleration due to gravity (-9.8m/s^2, assuming downward direction)
t is the time taken (which needs to be determined)

Rearranging the equation, we can solve for t:

25 = 0 + (1/2)(-9.8)t^2
t^2 = 25 / (0.5 * 9.8)
t^2 = 5.102
t ≈ √5.102
t ≈ 2.26 seconds

Therefore, it takes approximately 2.26 seconds for the stone to hit the cage.

2. Distance traveled by the cage before impact:
The distance traveled by the cage can be determined using the equation of motion:

s = ut + (1/2)at^2

where:
s is the displacement (which needs to be determined)
u is the initial velocity (0m/s for the cage as it starts from rest)
a is the acceleration (0.5m/s^2, assuming downward direction)
t is the time taken (which is equal to the time taken by the stone, 2.26 seconds)

Plugging in the values:

s = 0 + (1/2)(0.5)(2.26)^2
s = 0 + (1/2)(0.5)(5.102)
s ≈ 0.6376 meters

Therefore, the cage travels approximately 0.6376 meters before the stone hits it.

## distance the cage:

1/2 .5 t^2
distance the rock:
1/2 9.8 (t-t0) where to is found
25=1/2 *.5t^2 or t= 10 check that or
distance the rock:
1/2 9.8 (t-10)

now set the distances equal, solve for time of impact t.