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Chapter 2: Problem 65

Two students are on a balcony \(19.6 \mathrm{~m}\) above the street. One student throws a ball vertically downward at \(14.7 \mathrm{~m} / \mathrm{s} ;\) at the same instant, the other student throws a ball vertically upward at the same speed. The second ball just misses the balcony on the way down. (a) What is the difference in the two balls' time in the air? (b) What is the velocity of each ball as it strikes the ground? (c) How far apart are the balls \(0.800 \mathrm{~s}\) after they are thrown?

Short Answer

Expert verified
The time difference for the two balls in the air is about 1.00 s, the final velocity for the first ball is about 37.1 m/s and for the second ball is about -33.7 m/s, and the distance between the balls after 0.800 s is about 11.9 m.

Step by step solution

01

Determine the Time of Flight for the First Ball

Consider the first ball thrown downward with speed \(14.7 \mathrm{~m/s}\). Use the equation of motion \(h = ut + \frac{1}{2} gt^2\), where \(h = 19.6 \mathrm{~m}\) is the height, \(u = 14.7 \mathrm{~m/s}\) is the initial velocity, \(g = 9.8 \mathrm{~m/s^2}\) is acceleration due to gravity and \(t\) is the time. Solve this equation for \(t\).
02

Determine the Time of Flight for the Second Ball

Consider the second ball thrown upwards with speed \(14.7 \mathrm{~m/s}\). The ball will first go up then come back down. For the upward motion, use \(u = 14.7 \mathrm{~m/s}\), \(a = -9.8 \mathrm{~m/s^2}\) and \(h' = 0\) as the ball misses the balcony. You can get the time at the highest point from \(h' = ut - \frac{1}{2} gt'^2\). The total time of flight will be twice of \(t'\) as \(t' = t_{up} = t_{down}\).
03

Compute the Difference in Time

You can find the difference in time in the air by subtracting the time of flight of the first ball from that of the second.
04

Determine Velocity of Each Ball as it Strikes the Ground

Use the equation \(v = u + gt\) to determine the final velocities. For the first ball \(u = 14.7 \mathrm{~m/s}\) and for the second ball \(u = -14.7 \mathrm{~m/s}\), as it was thrown upward. Don't forget to use the respective individual times obtained in Steps 1 and 2 for each ball.
05

Determine Distance Between Balls after 0.800 s

After 0.800s, you can find the positions of both balls from the equation \(h = ut + \frac{1}{2} gt^2\). Repeat the calculations for both balls and subtract the two heights, keeping in mind the initial conditions and signs.

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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. In our exercise, kinematics helps us understand how and why the two balls move in different patterns. When analyzing a motion, it is crucial to understand:
  • Initial velocity: This is the speed at which an object starts moving. For instance, both balls here start with a speed of 14.7 m/s, but in different directions.
  • Displacement: This refers to the change in position from the starting point. Each ball travels a certain distance before hitting the ground.
  • Acceleration: It is the rate at which an object's velocity changes. Here, it is provided by gravity (9.8 m/s虏).
By understanding these aspects, we can predict how an object like a ball will travel through the air. Predicting the motion involves calculating how high it will go, how long it will stay up, or at what speed it will hit the ground.
Equations of Motion
Equations of motion are the tools that allow us to mathematically describe the movement of objects under the influence of forces. These equations relate displacement, initial velocity, final velocity, time, and acceleration. In this problem, we specifically use:
  • The basic equation for displacement:\[ h = ut + \frac{1}{2} gt^2 \]This equation helps find how far each ball travels depending on the time and the initial conditions.
  • The final velocity equation:\[ v = u + gt \]This one helps in calculating the speed at which each ball hits the ground.
Once you understand how to balance these equations with known and unknown variables, predicting motion in situations like our balcony exercise becomes manageable. Knowing which equation to apply based on the information given is key.
Acceleration Due to Gravity
Acceleration due to gravity is a natural phenomenon by which all objects are pulled toward the center of the Earth. In physics, this gravitational acceleration is expressed as approximately 9.8 m/s虏. This acceleration plays a crucial role in projectile motion problems. Regardless of whether an object is moving upwards or downwards, this force acts on it constantly:
  • It determines how quickly an object speeds up as it falls or slows down as it rises.
  • The 9.8 m/s虏 value ensures that all freely falling objects, regardless of their mass, accelerate towards Earth at the same rate in the absence of air resistance.
Understanding gravity's influence allows us to calculate both the peak heights and descent speeds for projectiles, ensuring accurate predictions in physics problems. In the case of the exercise, it determines the fate of the balls both as they rise and fall.

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Most popular questions from this chapter

It has been claimed that an insect called the froghopper (Philaenus spumarius) is the best jumper in the animal kingdom. This insect can accelerate at \(4000 \mathrm{~m} / \mathrm{s}^{2}\) over a distance of \(2.0 \mathrm{~mm}\) as it straightens its specially designed "jumping legs." (a) Assuming a uniform acceleration, what is the velocity of the insect after it has accelerated through this short distance, and how long did it take to reach that velocity? (b) How high would the insect jump if air resistance could be ignored? Note that the actual height obtained is about \(0.7 \mathrm{~m}\), so air resistance is important here.

A student throws a set of keys vertically upward to his fraternity brother, who is in a window \(4.00 \mathrm{~m}\) above. The brother's outstretched hand catches the keys \(1.50 \mathrm{~s}\) later. (a) With what initial velocity were the keys thrown? (b) What was the velocity of the keys just before they were caught?

A glider on an air track carries a flag of length \(\bar{\ell}\) through a stationary photogate, which measures the time interval \(\Delta t_{d}\) during which the flag blocks a beam of infrared light passing across the photogate. The ratio \(v_{d}=\) \(\ell / \Delta t_{d}\) is the average velocity of the glider over this part of its motion. Suppose the glider moves with constant acceleration. (a) Is \(v_{d}\) necessarily equal to the instantaneous velocity of the glider when it is halfway through the photogate in space? Explain. (b) Is \(v_{d}\) equal to the instantaneous velocity of the glider when it is halfway through the photogate in timc? Explain.

An ice sled powered by a rocket engine starts from rest on a large frozen lake and accelerates at \(+40 \mathrm{ft} / \mathrm{s}^{2}\). After some time \(t_{1}\), the rocket engine is shut down and the sled moves with constant velocity \(v\) for a time \(l_{2}\). If the total distance traveled by the sled is \(17500 \mathrm{ft}\) and the total time is \(90 \mathrm{~s}\), find (a) the times \(t_{1}\) and \(t_{2}\) and (b) the velocity \(\nu\). At the \(17500-\) ft mark, the sled begins to accelerate at \(-20 \mathrm{ft} / \mathrm{s}^{2}\), (c) What is the final position of the sled when it comes to rest? (d) How long does it take to come to rest?

A stuntman sitting on a tree limb wishes to drop vertically onto a horse galloping under the tree. The constant speed of the horse is \(10.0 \mathrm{~m} / \mathrm{s}\), and the man is initially \(3.00 \mathrm{~m}\) above the level of the saddle. (a) What must be the horizontal distance between the saddle and the limb when the man makes his move? (b) How long is he in the air?
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