A man stands on the edge of a cliff with hands outstretched horizontally
carrying two stones, one in each hand. At time
t= 0, he drops the first pebble from rest. After a time of 1.0 s, he releases the second pebble. At what time will the separation between the pebbles be 66 m?
To solve this problem, we need to consider the motion of both pebbles independently and determine when their separation reaches the desired value.
Let's start by analyzing the motion of each pebble individually.
Pebble 1:
- It is dropped from rest, so its initial velocity is 0 m/s.
- The time taken for pebble 1 to fall will be calculated using the equation: d = 0.5 * g * t^2, where d is the distance fallen, g is the acceleration due to gravity (approximately 9.8 m/s^2), and t is the time.
- Since the pebble falls for 1.0 s, substituting the given values into the equation, we find: 66 m = 0.5 * 9.8 m/s^2 * (1.0 s)^2.
Now, let's solve this equation to find the distance fallen by pebble 1.
66 m = 0.5 * 9.8 m/s^2 * (1.0 s)^2
66 m = 4.9 m/s^2 * (1.0 s)^2
66 m = 4.9 m * (1.0)^2
66 m = 4.9 m
From this equation, we observe that the distance fallen by pebble 1 is 4.9 m.
Now, let's analyze pebble 2:
- It is released after 1.0 s, so its time of flight will be 1.0 s shorter than pebble 1.
Since the separation between the pebbles is the distance fallen by pebble 1, we need to find the time taken by pebble 2 to fall the same distance.
Using the equation d = 0.5 * g * t^2:
4.9 m = 0.5 * 9.8 m/s^2 * t^2
Now, let's solve this equation to find the time taken by pebble 2.
4.9 m = 4.9 m/s^2 * t^2
t^2 = 1
t = 1 s
Therefore, pebble 2 takes 1 second to fall the same distance as pebble 1. Since pebble 2 is released after 1.0 s, the total time will be 1.0 s + 1.0 s = 2.0 s.
So, the separation between the pebbles will be 66 m after 2.0 seconds.
66=h1-h2=1/2 gt^2-1/2 g(t-1)^2
solve for time t.