A rock thrown from a bridge hits the water below at a horizontal distance x directly below the throwing point. The rock travels a smooth parabolic path in time t.

a. Show that the virtical distance of the bridge above water is 1/2gt^2.
b. What is the height of the bridge if the time the rock is airborne is 2 seconds?

To solve this problem, we can use the kinematic equations of motion. Let's break it down step by step:

a. To show that the vertical distance of the bridge above water is 1/2gt^2, we first need to consider the motion of the rock while in the air. We'll assume that the initial vertical velocity of the rock is zero when it is thrown horizontally.

Since the rock is thrown horizontally, we know its horizontal velocity remains constant. Let's call this velocity 'v' (which is equal to the distance x divided by the time t: v = x/t).

Now, let's consider the vertical motion of the rock. We know that the only force acting on the rock vertically is gravity. The acceleration due to gravity is denoted by 'g' and is equal to -9.8 m/s^2 (assuming Earth's gravity).

Using the kinematic equation:
h = v₀t + (1/2)gt^2

Since the initial vertical velocity of the rock is zero (v₀ = 0), the equation simplifies to:
h = (1/2)gt^2

Therefore, the vertical distance (h) of the bridge above water is 1/2gt^2.

b. Now, let's calculate the height of the bridge when the time the rock is airborne is 2 seconds.

Using the formula we derived in part a:
h = (1/2)gt^2

Substituting the given values:
h = (1/2)(-9.8 m/s^2)(2 s)^2
h = (1/2)(-9.8 m/s^2)(4 s^2)
h = (1/2)(-39.2 m^2/s^2)
h = -19.6 m^2/s^2

Since distance cannot be negative, we disregard the negative sign. The height of the bridge is 19.6 meters.

Therefore, the height of the bridge is 19.6 meters when the time the rock is airborne is 2 seconds.