Heather and Jerry are standing on a bridge 53m above a river. Heather throws a rock straight down with a speed of 12m/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?
a. wouldn't it just be the amount of time the rock thrown upward takes to pass the bridge on the way down?
To find the time elapsed between the first splash and the second splash, we need to determine the time it takes for each rock to reach the water.
Let's start by calculating the time it takes for Heather's rock to hit the water.
We are given that Heather throws the rock straight down with a speed of 12 m/s, and the height of the bridge is 53 m. We can use the equation of motion:
s = ut + (1/2)at^2
Where:
s = vertical distance traveled (53 m in this case)
u = initial velocity (12 m/s in this case)
a = acceleration (which is due to gravity and is approximately -9.8 m/s^2 for objects falling near the Earth's surface)
t = time
Plugging in the values:
53 = (12)t + (1/2)(-9.8)t^2
Simplifying the equation:
-4.9t^2 + 12t - 53 = 0
We can solve this quadratic equation for t using the quadratic formula:
t = (-b ± √(b^2 - 4ac)) / (2a)
In this case, a = -4.9, b = 12, and c = -53.
Calculating the values:
t = (-12 ± √(12^2 - 4(-4.9)(-53)) / (2*(-4.9))
Simplifying again:
t = (-12 ± √(144 - (-1029.2)) / (-9.8)
t = (-12 ± √(144 + 1029.2) / (-9.8)
t = (-12 ± √(1173.2) / (-9.8)
Now, we need to discard the negative value for t since time cannot be negative.
Calculating further:
t = (-12 + √(1173.2) / (-9.8)
t ≈ 2.856 seconds
Therefore, it takes approximately 2.856 seconds for Heather's rock to hit the water.
Since Jerry throws his rock at the exact same instant of time, the time elapsed between the first splash and the second splash is also approximately 2.856 seconds.