Problem 46 - College Physics

Question

A ball rolls horizontally with a speed of \(7.6 \mathrm{~m} / \mathrm{s}\) off the edge of a tall platform. If the ball lands \(8.7 \mathrm{~m}\) from the point on the ground directly below the edge of the platform, what is the height of the platform?

Step-by-Step Solution

Answer

The platform is approximately 6.4 meters tall.

Step 1: Analyzing Horizontal Motion

The ball rolls off the platform with a horizontal velocity. We are given the horizontal speed of the ball, \(v_x = 7.6 \text{ m/s}\), and the horizontal distance it travels is \(x = 8.7 \text{ m}\). Using the formula for distance, \(x = v_x \cdot t\), we find the time of flight: \[ t = \frac{x}{v_x} = \frac{8.7 \text{ m}}{7.6 \text{ m/s}} \approx 1.145 \text{ s} \]

Step 2: Analyzing Vertical Motion

The time of flight calculated in the horizontal motion is the same as the time it takes the ball to fall vertically. Since there is no initial vertical velocity and only gravity is acting on the ball in the vertical direction, we use:\[ y = \frac{1}{2}gt^2 \]where \(g = 9.8 \text{ m/s}^2\) is the acceleration due to gravity, and \(y\) is the height of the platform. Substitute the time and solve for \(y\):\[ y = \frac{1}{2} \times 9.8 \text{ m/s}^2 \times (1.145 \text{ s})^2 \approx 6.4 \text{ m}\]

Key Concepts

Horizontal Motion

When studying projectile motion, understanding horizontal motion is essential. In this scenario, a ball rolls off a platform with an initial horizontal velocity denoted as \( v_x = 7.6 \; \text{m/s} \). The horizontal motion is significant because it governs how far the ball travels along the ground. Here, the horizontal distance the ball covers is \( x = 8.7 \; \text{m} \).

To determine how long the ball is in the air, we use the formula for horizontal distance, \( x = v_x \times t \), to solve for time \( t \). This formula tells us that the time the ball takes to travel the horizontal distance is directly related to its horizontal speed.

So, we rearrange the formula to find \( t = \frac{x}{v_x} = \frac{8.7 \; \text{m}}{7.6 \; \text{m/s}}\), which gives us a time of approximately \(1.145 \) seconds. This calculated time duration is critical because it tells us how long gravity has to act on the ball.

Vertical Motion

Vertical motion in projectile problems deals with how objects fall under the influence of gravity, with no initial vertical velocity. This can be understood by dissecting how gravity affects an object in upward or downward motion. In this case, the ball dropped from the platform has no initial vertical velocity. Thus, gravity solely dictates its fall.

Here, the height \( y \) from which the ball falls is given by the equation \( y = \frac{1}{2}gt^2 \). This formula stems from the second law of motion, describing how distance changes with acceleration over time.

Since we want to find the height of the platform, substitute \( g = 9.8 \; \text{m/s}^2 \) for gravitational acceleration and \( t = 1.145 \; \text{seconds} \) as the time derived from horizontal motion. Thus, \( y = \frac{1}{2} \times 9.8 \; \text{m/s}^2 \times (1.145 \; \text{s})^2 \). Solving this will yield a height of approximately \( 6.4 \) meters, telling us how high the platform is from which the ball falls.

Acceleration due to Gravity

A fundamental force that acts on all objects in freefall is gravity. This constant force brings objects toward the Earth at a uniform acceleration, known as the acceleration due to gravity, denoted by \( g \).

Typically, \( g \) is valued at \( 9.8 \text{ m/s}^2 \), a crucial constant in physics as it directly affects vertical motion in projectile problems. It defines how quickly an object's velocity changes as it falls.

Understanding \( g \) helps explain why, regardless of the initial horizontal speed, the vertical drop is consistent for all objects once they are in freefall. This singular acceleration affects calculations where time is used to find vertical displacement, just as shown in how the height of a platform is computed.

• Gravitational acceleration does not depend on the mass of the falling object.
• It uniformly increases the velocity of freely falling objects.
• It provides the sole driving force for vertical motion in vacuums with no air resistance.