A rock is thrown horizontally from a bridge hits the water below 30 meter away in the horizontal direction. If the rock was in the air 2seconds, how tall is the bridge?

h = 0.5g*t^2 = 4.9*2^2 = 19.6 m.

To determine the height of the bridge, we can use the equations of motion, specifically the equation for vertical displacement. The formula for vertical displacement is given as:

Displacement (vertical) = Initial Vertical Velocity x Time + (0.5 x Acceleration x Time^2)

However, since the rock is thrown horizontally, and there is no vertical initial velocity and no vertical acceleration, the formula simplifies to:

Displacement (vertical) = 0.5 x Acceleration x Time^2

In this case, the rock is in the air for 2 seconds, so our equation becomes:

Displacement (vertical) = 0.5 x Acceleration x 2^2

Now, we need to find the acceleration. Since the rock is thrown horizontally, the horizontal and vertical motions are independent of each other. This means that the acceleration due to gravity only affects the vertical motion. The acceleration due to gravity is approximately 9.8 m/s^2.

Substituting the values into the equation, we have:

Displacement (vertical) = 0.5 x 9.8 x 2^2
= 0.5 x 9.8 x 4
= 19.6 meters

Therefore, the height of the bridge is approximately 19.6 meters.