A rock thrown horizontally 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 Therapy vertical distance of the bridge above water is 1/2gr2.
B.) What is the height of the bridge if the time of the rock is airborne is 2 seconds?

How can the rock land "directly below the throwing point" and a horizontal distance x away?

Perhaps you left out some important words of the question.

A.) To determine the vertical distance of the bridge above the water, we can start by analyzing the horizontal motion of the rock. Since the rock is thrown horizontally, there is no initial vertical velocity (vy = 0 m/s). However, there is a constant acceleration due to gravity acting vertically downward (ay = -g m/s^2).

During the time t, the rock will cover a horizontal distance x. The horizontal distance traveled is given by the equation:

x = vx * t

Since the rock is thrown horizontally, the horizontal velocity (vx) remains constant throughout the motion. Thus, we can rewrite the equation as:

x = v * t ............(1)

In the vertical direction, the initial vertical position (y0) is equal to zero since the rock is thrown from the bridge. The vertical displacement (Δy) can be determined using the equation:

Δy = v0y * t + 0.5 * a * t^2

Since the initial vertical velocity (v0y) is zero, the equation simplifies to:

Δy = 0.5 * a * t^2 ............(2)

Now, we need to relate the horizontal distance (x) with the vertical displacement (Δy). Since the rock travels along a smooth parabolic path, we know that the horizontal distance x is equal to twice the horizontal component of the initial velocity (v0x) multiplied by the time t:

x = 2 * v0x * t

Since the rock is thrown horizontally, the initial vertical velocity (v0y) is zero. Therefore, the initial velocity (v0) is equal to the horizontal velocity (vx):

v0 = vx

Thus, we can substitute v for v0 in equation (1) and rewrite it as:

x = v * t ............(3)

We can equate equations (2) and (3) to eliminate the variable t:

0.5 * a * t^2 = v * t

Simplifying the equation by canceling the common factor t, we get:

0.5 * a * t = v

Rearranging the equation for v, we find:

v = 0.5 * a * t ............(4)

Substituting equation (4) into equation (3) to eliminate v and x, we get:

0.5 * a * t = x

Simplifying the equation by canceling the common factor 0.5, we find:

a * t = 2x

Rearranging the equation for t, we find:

t = (2x) / a ............(5)

Now, substituting equation (5) into equation (2) to eliminate t, we find:

Δy = 0.5 * a * t^2

Δy = 0.5 * a * [(2x) / a]^2

Δy = 0.5 * a * (4x^2 / a^2)

Simplifying the equation by canceling the common factors, we get:

Δy = 2x^2 / a

Thus, the vertical distance of the bridge above the water is given by:

Δy = 2x^2 / a

Substituting the acceleration due to gravity (g) for a, we find:

Δy = 2x^2 / g

B.) To determine the height of the bridge if the time of the rock's airborne motion is 2 seconds, we can use the equation derived in part A:

Δy = 2x^2 / g

Since the rock is thrown horizontally, the horizontal distance x is equal to the horizontal component of the initial velocity (v0x) multiplied by the time t:

x = v0x * t

Since the rock is thrown horizontally, the initial vertical velocity (v0y) is zero. Therefore, the initial velocity (v0) is equal to the horizontal velocity (vx):

v0 = vx

Thus, we can substitute v for v0 in the equation for x:

x = v * t

Substituting this value of x into the equation for Δy, we get:

Δy = 2(v * t)^2 / g

Since the height of the bridge is the vertical distance above the water, we can substitute Δy with h (height of the bridge):

h = 2(v * t)^2 / g

Now, we need to find the value of v (velocity) and time t. Given that the time of the rock's airborne motion is 2 seconds, we can substitute this value into the equation:

h = 2(v * 2)^2 / g

Simplifying the equation, we get:

h = 8(v^2) / g

To determine the height of the bridge, we need to know the magnitude of the velocity (v) with which the rock is thrown.

To solve this problem, we need to use the equations of motion and the principles of projectile motion. Let's break it down step by step:

A.) To find the vertical distance of the bridge above the water (h), we need to find the height at which the rock is thrown (h₀). The rock is in free fall vertically, so we can use the equation for the height of an object in free fall:

h₀ = 1/2 * g * t²

Here, g is the acceleration due to gravity (approximately 9.8 m/s²), and t is the time of flight of the rock.

Now, we need to find the horizontal distance (x) covered by the rock. Since the rock is thrown horizontally, it will have a horizontal velocity (V₀x), but no vertical velocity.

x = V₀x * t

Since the rock is thrown horizontally, the horizontal velocity remains constant throughout the motion. Therefore, we can relate the horizontal velocity V₀x to the initial speed (V₀) of the rock using the equation:

V₀x = V₀ * cos(θ)

Where θ is the angle at which the rock is thrown. Since the rock is thrown horizontally, the angle θ is 0 degrees, so cos(θ) = 1.

x = V₀ * t

Now, we can substitute the equation for x into the equation for h₀:

h₀ = 1/2 * g * (x / V₀)²

Simplifying this equation further:

h₀ = 1/2 * g * (V₀ * t / V₀)²
h₀ = 1/2 * g * t²

Therefore, the vertical distance of the bridge above the water is equal to 1/2 * g * t².

B.) Now, if the time of flight of the rock is 2 seconds (t = 2), we can use the equation derived in part A to find the height of the bridge (h).

h = 1/2 * g * t²
h = 1/2 * 9.8 * (2)²
h = 19.6 meters

Therefore, the height of the bridge is 19.6 meters if the time of the rock's airborne is 2 seconds.