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A ball is thrown straight up from the ground with speed \(v_{0} .\) At the same instant, a second ball is dropped from rest from a height \(H,\) directly above the point where the first ball was thrown upward. There is no air resistance. (a) Find the time at which the two balls collide. (b) Find the value of \(H\) in terms of \(v_{0}\) and \(g\) so that at the instant when the balls collide, the first ball is at the highest point of its motion.

Short Answer

Expert verified
(a) The balls collide at time \(t = \frac{H}{v_0}\). (b) Height \(H = \frac{v_0^2}{g}\).

Step by step solution

01

Establish Equations of Motion

The first ball is thrown upwards with an initial velocity \(v_0\). Its position as a function of time \(t\) is given by \(y_1(t) = v_0 t - \frac{1}{2}gt^2\). The second ball is dropped from a height \(H\) and its position as a function of time \(t\) is \(y_2(t) = H - \frac{1}{2}gt^2\).
02

Set Up the Collision Condition

For the balls to collide, their positions at time \(t\) must be the same: \(y_1(t) = y_2(t)\). Set the equations equal to each other: \(v_0 t - \frac{1}{2}gt^2 = H - \frac{1}{2}gt^2\).
03

Cancel Terms and Solve for Time

Notice that \(-\frac{1}{2}gt^2\) cancels on both sides, so you're left with \(v_0 t = H\). Solving for \(t\) gives \(t = \frac{H}{v_0}\). This is the time at which the balls collide.
04

Condition for the Highest Point of Motion

At the highest point of the first ball's motion, its velocity becomes zero. Using the equation \(v = v_0 - gt\), setting \(v = 0\), we find that the time to reach the highest point is \(t = \frac{v_0}{g}\). For the collision to happen at this point, set \(\frac{H}{v_0} = \frac{v_0}{g}\).
05

Solve for Height H

From the equation \(\frac{H}{v_0} = \frac{v_0}{g}\), solve for \(H\): \(H = \frac{v_0^2}{g}\). Thus, \(H\) in terms of \(v_0\) and \(g\) is \(\frac{v_0^2}{g}\).

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Key Concepts

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

Kinematics
Kinematics is the branch of physics that deals with the motion of objects without considering the causes of this motion. It provides a way to describe the positions, velocities, and accelerations of objects using mathematical equations, typically in terms of time.

In the given problem, kinematics helps us understand how two balls move over time. The first ball is thrown upwards, fighting against gravity, while the second ball is dropped downwards. Both of these trajectories can be described using kinematic equations. This makes it possible to predict where the balls will be at any given moment.

By analyzing their motion, we can also solve for conditions like when and where they might collide, which is a typical kinematic problem. Remember, these calculations often ignore other forces, like air resistance, to simplify the analysis.
Equations of Motion
Equations of motion are fundamental to solving kinematic problems. They are mathematical formulas that help predict an object's future position and velocity based on its initial conditions.

For the problem at hand, the position of the first ball can be expressed as:
  • \( y_1(t) = v_0 t - \frac{1}{2}gt^2 \), where \( v_0 \) is the initial speed and \( g \) is gravitational acceleration.
Similarly, the position of the second ball is:
  • \( y_2(t) = H - \frac{1}{2}gt^2 \), starting from height \( H \).
These equations form the basis for calculating the time and height when the balls meet. By setting the two position equations equal, we can find when their trajectories collide.
Collision Condition
The collision condition refers to a scenario where two or more moving objects meet at the same place at the same time. In our exercise, the collision occurs when both balls have the same vertical position.

Given the position equations:
  • \( y_1(t) = y_2(t) \) — this sets the stage for solving where and when the two positions become identical.
This leads us to the equation:
  • \( v_0 t = H \).
By solving \( t = \frac{H}{v_0} \), we determine the time at which their paths converge. This ensures that you can predict precisely when and where two paths intersect, using their motion laws.
Gravitational Acceleration
Gravitational acceleration is the constant rate at which objects accelerate towards the Earth due to gravity, commonly denoted as \( g \). It's approximately \( 9.81 \, \text{m/s}^2 \) near the Earth's surface.

This constant plays a big role in determining how objects move in projectile motion, especially when they are only influenced by gravity. It affects how quickly the balls from the example rise or fall over time.

By understanding gravitational acceleration, you can predict how an object's velocity changes over time. In the example problem, \( g \) is critical to computing the position of both balls and ultimately solving for the meeting point in their trajectories.

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Most popular questions from this chapter

The fastest measured pitched baseball left the pitcher's hand at a speed of 45.0 \(\mathrm{m} / \mathrm{s}\) . If the pitcher was in contact with the ball over a distance of 1.50 \(\mathrm{m}\) and produced constant acceleration, (a) what acceleration did he give the ball, and (b) how much time did it take him to pitch it?

A helicopter carrying Dr. Evil takes off with a constant upward acceleration of 5.0 \(\mathrm{m} / \mathrm{s}^{2}\) . Secret agent Austin Powers jumps on just as the helicopter lifts off the ground. After the two men struggle for 10.0 s, Powers shuts off the engine and steps out of the helicopter. Assume that the helicopter is in free fall after its engine is shut off, and ignore the effects of air resistance. (a) What is the maximum height above ground reached by the helicopter? (b) Powers deploys a jet pack strapped on his back 7.0 s after leaving the helicopter, and then he has a constant downward acceleration with magnitude 2.0 \(\mathrm{m} / \mathrm{s}^{2} .\) How far is Powers above the ground when the helicopter crashes into the ground?

CALC A car is stopped at a traffic light. It then travels along a straight road so that its distance from the light is given by \(x(t)=b t^{2}-c t^{3},\) where \(b=2.40 \mathrm{m} / \mathrm{s}^{2}\) and \(c=0.120 \mathrm{m} / \mathrm{s}^{3} .\) (a) Calculate the average velocity of the car for the time interval \(t=0\) to \(t=10.0\) s. (b) Calculate the instantaneous velocity of the car at \(t=0, t=5.0 \mathrm{s},\) and \(t=10.0 \mathrm{s}\) . (c) How long after starting from rest is the car again at rest?

A 15-kg rock is dropped from rest on the earth and reaches the ground in 1.75 \(\mathrm{s}.\) When it is dropped from the same height on Saturn's satellite Enceladus, it reaches the ground in 18.6 \(\mathrm{s}.\) What is the acceleration due to gravity on Enceladus?

The position of the front bumper of a test car under microprocessor control is given by \(x(t)=2.17 \mathrm{m}+\) \(\left(4.80 \mathrm{m} / \mathrm{s}^{2}\right) t^{2}-\left(0.100 \mathrm{m} / \mathrm{s}^{6}\right) t^{6} .\) (a) Find its position and acceleration at the instants when the car has zero velocity. (b) Draw \(x\) -t, \(v_{x}-t,\) and \(a_{x}-t\) graphs for the motion of the bumper between \(t=0\) and \(t=2.00 \mathrm{s}.\)

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