A rock is thrown off a bridge at an angle of 38degree above horizontal, and landed in the water 3.60 seconds later 60.0 meters from the point at the waters level directly below the thrower... What is the rock's horizontal velocity?

I need better clarity on physics, please?

you have distance, time

horizonal velocity=horizontaldistance/time
= 60/3.6 m/s

Sure! In this problem, we have a rock being thrown off a bridge at an angle of 38 degrees above the horizontal. We need to find the rock's horizontal velocity.

To start solving this problem, we can break down the velocity of the rock into its horizontal and vertical components. The horizontal velocity (Vx) remains the same throughout the motion, while the vertical velocity (Vy) changes due to the effect of gravity.

Let's use the following equations of motion to solve the problem:

Vertical Motion:
Vy = V0y + g * t
Where:
- Vy is the final vertical velocity
- V0y is the initial vertical velocity
- g is the acceleration due to gravity (approximately 9.8 m/s^2)
- t is the time taken for the rock to reach the water

Horizontal Motion:
Sx = Vx * t
Where:
- Sx is the horizontal distance covered by the rock
- Vx is the horizontal velocity
- t is the time taken for the rock to reach the water

By analyzing the problem, we can determine that the initial vertical velocity (V0y) is equal to the initial velocity of the rock. Since the rock is thrown off at an angle of 38 degrees above the horizontal, we can calculate V0y using the formula:

V0y = V0 * sin(theta)
Where:
- V0y is the initial vertical velocity
- V0 is the initial velocity of the rock (which we need to find)
- theta is the angle at which the rock is thrown (38 degrees)

Once we have calculated V0y, we can use it in the first equation to find Vy at the moment the rock hits the water.

Now, the horizontal distance covered by the rock is given as 60.0 meters. We are also given the time taken for the rock to reach the water, which is 3.60 seconds.

Using the second equation for horizontal motion, we can rearrange it to solve for Vx:

Vx = Sx / t
Where:
- Vx is the horizontal velocity
- Sx is the horizontal distance covered by the rock (60.0 meters)
- t is the time taken for the rock to reach the water (3.60 seconds)

By substituting the given values into the equation, we can find the horizontal velocity of the rock.

I hope this clarifies how to approach the physics problem! Let me know if there's anything else I can help you with.