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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?

Short Answer

Expert verified
The platform is approximately 6.4 meters tall.

Step by step solution

01

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: \[ t = \frac{x}{v_x} = \frac{8.7 \text{ m}}{7.6 \text{ m/s}} \approx 1.145 \text{ s} \]
02

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 \(y\):\[ y = \frac{1}{2} \times 9.8 \text{ m/s}^2 \times (1.145 \text{ s})^2 \approx 6.4 \text{ m}\]

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

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.

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Most popular questions from this chapter

A ball rolling on a table has a velocity with rectangular components \(v_{x}=0.60 \mathrm{~m} / \mathrm{s}\) and \(v_{y}=0.80 \mathrm{~m} / \mathrm{s} .\) What is the displacement of the ball in an interval of \(2.5 \mathrm{~s} ?\)

A person riding in the back of a pickup truck traveling at \(70 \mathrm{~km} / \mathrm{h}\) on a straight, level road throws a ball with a speed of \(15 \mathrm{~km} / \mathrm{h}\) relative to the truck in the direction opposite to the truck's motion. What is the velocity of the ball (a) relative to a stationary observer by the side of the road, and (b) relative to the driver of a car moving in the same direction as the truck at a speed of \(90 \mathrm{~km} / \mathrm{h} ?\)

A hockey puck slides along a horizontal ice surface at \(20.0 \mathrm{~m} / \mathrm{s}\), hits a flat vertical wall, and bounces off. Its initial velocity vector makes an angle of \(35^{\circ}\) with the wall and it comes off at an angle of \(25^{\circ}\) moving at \(10.0 \mathrm{~m} / \mathrm{s}\). Choose the \(+x\) -axis to be along the wall in the direction of motion and the \(y\) -axis to be perpendicular (into) to the wall. (a) Write each velocity in unit vector notation. (b) Determine the change in velocity in unit vector notation. (c) Determine the magnitude and direction, relative to the wall, of this velocity change.

You are in a fast powerboat that is capable of a sustained steady speed of \(20.0 \mathrm{~m} / \mathrm{s}\) in still water. On a swift, straight section of a river you travel parallel to the bank of the river. You note that you take \(15.0 \mathrm{~s}\) to go between two trees on the riverbank that are \(400 \mathrm{~m}\) apart. (a) (1) Are you traveling with the current, (2) are you traveling against the current, or (3) is there no current? (b) If there is a current [reasoned in part (a)], determine its speed.

An observer by the side of a straight, level, northsouth road watches a car (A) moving south at a rate of \(75 \mathrm{~km} / \mathrm{h}\). A driver in another car (B) going north at \(50 \mathrm{~km} / \mathrm{h}\) also observes car \(\mathrm{A}\). (a) What is car A's velocity as observed from car B? (Take north to be positive.) (b) If the roadside observer sees car A brake to a stop in \(6.0 \mathrm{~s}\) what constant acceleration would be measured? (c) What constant acceleration would the driver in car B measure for the braking car A?

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