A stone is dropped from the roof of a high building. A second stone is dropped 1.08s later. How far apart are the stone when the second one has reached a speed of 15.0?
2nd:
t2 = 15/9.8
x2 = 1/2 (9.8) t2^2
1st:
t1 = t2+1.08
x1= 1/2(9.8)t1^2
Subtract x1-x2 to find out how far apart
To calculate the distance between the two stones when the second one has reached a speed of 15.0 m/s, we need to determine the time it took for the second stone to reach that speed.
First, we know that the first stone was dropped from the roof, so its initial speed is 0 m/s. From the information given, we can assume that the acceleration due to gravity is approximately 9.8 m/s².
Since we are interested in the time it takes for the second stone to reach a speed of 15.0 m/s, we can use the formula for the final velocity of an object in free fall:
v = u + at,
where
v is the final velocity,
u is the initial velocity,
a is the acceleration, and
t is the time taken.
For the first stone, we have:
v1 = 15.0 m/s (final velocity),
u1 = 0 m/s (initial velocity),
a = 9.8 m/s² (acceleration due to gravity).
Rearranging the equation, we get:
t1 = (v1 - u1) / a.
Substituting the values, we have:
t1 = (15.0 m/s - 0 m/s) / 9.8 m/s² = 1.53 s.
To find the distance between the two stones, we need to calculate the distance the first stone traveled in 1.08 s. We know that the distance traveled by an object in free fall is given by the equation:
d = ut + 1/2 * at²,
where
d is the distance,
u is the initial velocity,
t is the time taken, and
a is the acceleration.
For the first stone, we have:
d = 0.5 * 9.8 m/s² * (1.08 s)² = 5.32 m.
So, when the second stone reaches a speed of 15.0 m/s, it will be (5.32 m - 0 m) = 5.32 m vertically below the first stone.
Therefore, the two stones will be 5.32 meters apart vertically when the second stone reaches a speed of 15.0 m/s.