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A ball is thrown upward from the ground with an initial speed of \(25 \mathrm{~m} / \mathrm{s}\); at the same instant, another ball is dropped from a building \(15 \mathrm{~m}\) high. After how long will the balls be at the same height?

Short Answer

Expert verified
The balls will be at the same height after 0.6 seconds.

Step by step solution

01

Define the Variables and Equations

Let's denote the height of the first ball (thrown upward) as \(h1\) and the height of the second ball (dropped from a building) as \(h2\). For the first ball, we have \(h1 = h_{0,1} + v_{0,1} * t - 0.5 * g * t^2\), where \(h_{0,1}\) is the initial height, \(v_{0,1}\) is the initial velocity and \(g\) is the gravitational acceleration. For the second ball, we have \(h2 = h_{0,2} - 0.5 * g * t^2\), where \(h_{0,2}\) is the initial height of the second ball. The task is to find the time \(t\) when \(h1 = h2\).
02

Substituting Given Values

Plug the values into the respective equations: for the first ball, \(h_{0,1} = 0 m\), \(v_{0,1} = 25 m/s\), \(g = 9.81 m/s^2\); for the second ball, \(h_{0,2} = 15 m\), \(g = 9.81 m/s^2\). This gives us two equations: \(h1 = 0 m + 25 m/s * t - 0.5 * 9.81 m/s^2 * t^2\) and \(h2 = 15 m - 0.5 * 9.81 m/s^2 * t^2\).
03

Solve The Equations

Setting the equations \(h1\) and \(h2\) equal to each other gives: \(0 m + 25 m/s * t - 0.5 * 9.81 m/s^2 * t^2 = 15 m - 0.5 * 9.81 m/s^2 * t^2\). After simplifying, we get the equation \(25 m/s * t - 15 m = 0\) from which we solve for \(t\) to get \(t = 15 m / (25 m/s) = 0.6 s\). This is the time when the balls will be at the same height.

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Key Concepts

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

Kinematics
Kinematics is a branch of physics that studies the motion of objects without considering the forces that cause this motion. In the problem discussed, kinematics helps to analyze the motion of the two balls. One ball is thrown upwards, and another is dropped downwards. By understanding kinematics, we can predict and calculate how each ball moves over time. This involves using parameters such as initial velocity, height, time, and acceleration.

The motion of the balls is described using kinematic equations, which include initial velocities and gravitational effects. For example:
  • Initial speed: It determines how fast an object starts moving.
  • Acceleration: Here, it is due to gravity, which constantly affects the speed.
  • Time: The variable that indicates how long the ball has been moving.
In this situation, both balls follow a parabolic path because of the uniform gravitational pull acting on them, exemplifying typical kinematic motion.
Gravitational Acceleration
Gravitational acceleration is the rate at which an object increases its velocity as it falls towards the Earth due to gravity. It is denoted by the symbol \( g \) and is approximately \( 9.81 \ m/s^2 \) near the Earth's surface.

This concept is crucial in projectile motion as it influences how quickly objects fall and how trajectories are shaped. The ball thrown upward in the problem decelerates until it stops momentarily at its peak height, then accelerates downwards due to gravity. Meanwhile, the other ball, dropped from a height, accelerates from the start.

Understanding gravitational acceleration allows us to:
  • Predict the motion of falling objects
  • Calculate key values like time of flight
  • Determine maximum height reached
Hence, gravity not only affects how fast an object falls but also the entire trajectory and time of motion in projectile scenarios.
Equations of Motion
In the context of projectile motion, equations of motion are mathematical expressions that describe the movement of objects under specific conditions like initial velocity and acceleration.

The three standard kinematic equations in physics often involve parameters like displacement (\( h \)), initial velocity (\( v_0 \)), time (\( t \)), and acceleration due to gravity (\( g \)). Each equation correlates with different terms to provide insight into different motion aspects:
  • Displacement equation: \( h = h_0 + v_0 \cdot t - 0.5 \cdot g \cdot t^2 \)
  • Velocity equation: \( v = v_0 + g \cdot t \)
  • Acceleration consideration
For the exercise, these equations help us express the relation between the two balls' positions over time and solve for when their heights equal. Subsequently, setting up and rearranging these equations enables solving for time when both balls meet at the same height.

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Most popular questions from this chapter

A car traveling east at \(40.0 \mathrm{~m} / \mathrm{s}\) passes a trooper hiding at the roadside. The driver uniformly reduces his speed to \(25.0 \mathrm{~m} / \mathrm{s}\) in \(3.50 \mathrm{~s}\). (a) What is the magnitude and direction of the car's acceleration as it slows down? (b) How far does the car travel in the \(3.5-\mathrm{s}\) time period?

A small mailbag is released from a helicopter that is descending steadily at \(1.50 \mathrm{~m} / \mathrm{s}\). After \(2.00 \mathrm{~s}\), (a) what is the speed of the mailbag, and (b) how far is it below the helicopter? (c) What are your answers to parts (a) and (b) if the helicopter is rising steadily at \(1.50 \mathrm{~m} / \mathrm{s} ?\)

A hockey player is standing on his skates on a frozen pond when an opposing player, moving with a uniform speed of \(12 \mathrm{~m} / \mathrm{s}\), skates by with the puck. After \(3.0 \mathrm{~s}\), the first player makes up his mind to chase his opponent. If he accelerates uniformly at \(4.0 \mathrm{~m} / \mathrm{s}^{2}\), (a) how long does it take him to catch his opponent, and (b) how far has he traveled in that time? (Assume the player with the puck remains in motion at constant speed.)

A truck covers \(40.0 \mathrm{~m}\) in \(8.50 \mathrm{~s}\) while smoothly slowing down to a final velocity of \(2.80 \mathrm{~m} / \mathrm{s}\). (a) Find the truck's original speed. (b) Find its acceleration.

An attacker at the base of a castle wall \(3.65 \mathrm{~m}\) high throws a rock straight up with speed \(7.40 \mathrm{~m} / \mathrm{s}\) at a height of \(1.55 \mathrm{~m}\) above the ground. (a) Will the rock reach the top of the wall? (b) If so, what is the rock's speed at the top? If not, what initial speed must the rock have to reach the top? (c) Find the change in the speed of a rock thrown straight down from the top of the wall at an initial speed of \(7.40 \mathrm{~m} / \mathrm{s}\) and moving between the same two points. (d) Does the change in speed of the downward-moving rock agree with the magnitude of the speed change of the rock moving upward between the same elevations? Explain physically why or why not.

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