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?
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.
Wrong the answers are 0.4 seconds or 4 seconds
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.