Question

You are an imagineer and need to make sure that the second rock falls without actually hitting the actor.
The first rock has a mass of 2 1/2 (2.5) kilograms.
From a height of 78.4 meters, it hit the ground in 4 seconds.
Your second rock is 25 kilograms. (10 times the mass of the first rock).
From the same height, how long will it take for the SECOND rock to fall?

Answer 1

Wrong the answers are 0.4 seconds or 4 seconds

Answer 2

To find the time it will take for the second rock to fall, we can use the equation for free fall:

h = (1/2)gt^2

Where:
h = height (78.4 meters)
g = acceleration due to gravity (9.8 m/s^2)
t = time

First, we can calculate the time it took for the first rock to fall:

78.4 = (1/2)(9.8)t^2
78.4 = 4.9t^2
t^2 = 16
t = 4 seconds

Now we can use the time it took for the first rock to fall to calculate the time it will take for the second rock to fall:

78.4 = (1/2)(9.8)t^2
78.4 = 4.9t^2
t^2 = 16(10)
t = 40 seconds

Therefore, it will take the second rock 40 seconds to fall from the same height of 78.4 meters.

Answer 3

Apologies for the error. Let's recalculate:

Given:
- First rock: m1 = 2.5 kg, t1 = 4 seconds
- Second rock: m2 = 25 kg, h = 78.4 meters

Using the equation for free fall:

h = (1/2)gt^2

For the first rock:
78.4 = (1/2)(9.8)t1^2
78.4 = 4.9t1^2
t1^2 = 16
t1 = 4 seconds

Now, let's calculate the time it will take for the second rock to fall:

78.4 = (1/2)(9.8)t2^2
78.4 = 4.9t2^2
t2^2 = 16
t2 = 4 seconds

Therefore, the time it will take for the second rock to fall from the same height of 78.4 meters is also 4 seconds.