Heather and Jerry are standing on a bridge 51 m above a river. Heather throws a rock straight down with a speed of 17 m/s . Jerry, at exactly the same instant of time, throws a rock straight up with the same speed. Ignore air resistance.
How much time elapses between the first splash and the second splash?
h=vi*t-4.9t^2
solve for t, given vi=+17, then vi=-17
h=-51
To calculate the time elapsed between the first splash and the second splash, we can use the concept of free fall and the kinematic equation for falling objects.
First, let's calculate the time it takes for the rock thrown by Heather to hit the river. We can use the equation:
h = (1/2)gt^2
where h is the height, g is the acceleration due to gravity (approximately 9.8 m/s^2), and t is the time.
Rearranging the equation to solve for time, we have:
t = √(2h/g)
Substituting the given values, we have:
t = √(2 * 51 m / 9.8 m/s^2)
= √(102 / 9.8)
≈ √10.41
≈ 3.23 s
Therefore, it takes approximately 3.23 seconds for the rock thrown by Heather to hit the river.
Since Jerry throws the rock straight up with the same initial speed, the time it takes for the rock to reach its highest point is equal to the time it takes for the rock to hit the river.
Therefore, the time elapsed between the first splash and the second splash is also approximately 3.23 seconds.
To determine the amount of time that elapses between the first splash and the second splash, we need to find the time it takes for each rock to reach the river.
We can use the kinematic equation for vertical motion to find the time of flight for each rock.
For Heather's rock:
We know the initial velocity is 17 m/s, the final velocity is 0 m/s (when it reaches the river), the acceleration is due to gravity (-9.8 m/s^2), and the displacement is -51 m (negative because it is directed downward).
Using the equation v = u + at, where:
v = final velocity, u = initial velocity, a = acceleration, and t = time,
we can substitute the known values into the equation:
0 = 17 + (-9.8)t
Solving for t, we get:
-9.8t = -17
t = 17/9.8 ≈ 1.7347 seconds
Jerry's rock:
Since Jerry throws his rock straight up with the same speed, the initial velocity is 17 m/s, but the final velocity is now negative (-17 m/s) because it will have a downward velocity when it reaches the river. The displacement is still -51 m.
Using the same equation as before:
v = u + at
-17 = 17 + (-9.8)t
Solving for t, we get:
-9.8t = -34
t = 34/9.8 ≈ 3.4694 seconds
So the time difference between the first splash and the second splash is the difference between the two times:
3.4694 - 1.7347 ≈ 1.7347 seconds
Therefore, approximately 1.7347 seconds elapse between the first splash and the second splash.