91Ó°ÊÓ

In projectile motion, horizontal motion occurs independently from vertical motion. The key thing to remember is that the horizontal velocity remains constant throughout the motion unless acted upon by an external force. For this exercise, let's review the basic concept using the ball rolling off the stairway.

Since there is no horizontal acceleration, we use the formula:

\( x = v_x t\)

Here:

• \( x \) is the horizontal distance the ball travels.
• \(v_x \) is the constant horizontal speed of the ball (1.52 m/s in this case).
• \( t \) is the time elapsed.

This equation tells us how far the ball will travel horizontally in a given time period. Let's remember that time (\( t \) ) plays a crucial role in determining how far the ball goes both horizontally and vertically.

Vertical motion, unlike horizontal motion, is influenced by gravity. Gravity causes a constant acceleration downwards. Hence, the vertical distance ( \( y \) ) the ball travels is given by:
\( y = \frac{1}{2} g t^2 \)

Where:

• \( y \) is the vertical distance.
• \( g \) is the acceleration due to gravity (approx. 9.8 m/s²).
• \( t \) is the time elapsed.

Using the given problem, the ball falls and covers a certain vertical distance by the time it hits the step. This motion is combined with the horizontal motion, and both are dependent on the same \( t \).

It is important to express the time from the horizontal motion equation, which is: \( t = \frac{x}{v_x} \). By substituting this into the vertical motion equation, we tie both motions together explicitly.
This creates the unified equation:
\( y = \frac{1}{2} g \left( \frac{x}{v_x} \right)^2 \).
This equation shows how far the ball falls vertically based on how far it’s traveled horizontally.

Equations of motion are vital for analyzing projectile motion problems, helping you connect horizontal and vertical motion components.
Let's look at the equations used in the problem again.

For horizontal motion:

• \( x = v_x t \)

For vertical motion:

• \( y = \frac{1}{2} g t^2 \)

When combining these equations, time becomes the link between the two motions.


By expressing the time from the horizontal motion:
\( t = \frac{x}{v_x} \).
This expression helps relating the two motions directly:
\( y = \frac{1}{2} g \left( \frac{x}{v_x} \right)^2 \).

In our exercise, we calculated where the ball hits the step using this unified equation. For the ball to hit the n-th step, the vertical distance covered should equal the step height (0.203 m):

\( y = n \times 0.203 \).
By substituting this into our unified equation and solving for n, we find out which step the ball hits first.
Upon solving: \( n ≈ 3 \).
Thus, the ball hits the third step first. Understanding these equations can help better visualize and solve similar projectile motion problems.