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A student throws a water balloon vertically downward from the top of a building. The balloon leaves the thrower's hand with a speed of 6.00 \(\mathrm{m} / \mathrm{s}\) . Air resistance may be ignored, so the water balloon is in free fall after it leaves the thrower's hand. (a) What is its speed after falling for 2.00 s? (b) How far does it fall in 2.00 \(\mathrm{s?}\) (c) What is the magnitude of its velocity after falling 10.0 \(\mathrm{m} ?\) (d) Sketch \(a_{y}-t, v_{y}-t,\) and \(y-t\) graphs for the motion.

Short Answer

Expert verified
(a) 25.62 m/s (b) 21.62 m (c) 15.19 m/s.

Step by step solution

01

Understand the Problem

The water balloon is thrown downward from a building. It's important to note that after it's released, it only has two forces acting on it: gravity pulling it downward and no air resistance affecting it. We will calculate the speed after some time has passed and how far it falls.
02

Identify Known Values

The initial velocity ( v_0 ) is 6.00 m/s downward. The gravitational acceleration ( g ) is approximately 9.81 m/s². We need to analyze the situation for times ranging from 2.00 seconds, and for a fall of 10.0 meters.
03

Calculate Speed After 2 Seconds

Use the formula for velocity in free fall: v = v_0 + gt . Substitute v_0 = 6.00 m/s, g = 9.81 m/s², and t = 2.00 seconds:\[ v = 6.00 + (9.81)(2.00) = 25.62 \] m/s.
04

Find Distance Fallen in 2 Seconds

Use the formula for distance in free fall: y = v_0t + \frac{1}{2}gt^2 . Substitute v_0 = 6.00 m/s, g = 9.81 m/s², and t = 2.00 seconds:\[ y = (6.00)(2.00) + \frac{1}{2}(9.81)(2.00)^2 = 21.62 \] meters.
05

Determine Velocity After Falling 10 Meters

Use the kinematic equation: v^2 = v_0^2 + 2gy . Substitute v_0 = 6.00 m/s, g = 9.81 m/s², and y = 10.0 meters:\[ v^2 = 6.00^2 + 2(9.81)(10.0) \]\[ v = \sqrt{36.00 + 196.2} = 15.19 \] m/s.
06

Sketch Graphs for Motion

1. Acceleration vs. Time ( a_y - t graph): This will be a horizontal line at 9.81 m/s² since acceleration due to gravity is constant. 2. Velocity vs. Time ( v_y - t graph): This will be a straight line starting from 6.00 m/s and increasing linearly. 3. Position vs. Time ( y - t graph): This will be a parabolic curve opening upward, representing increasing distance with time.

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Key Concepts

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

Free Fall
Free fall occurs when an object is moving under the sole influence of gravity. This means that no other forces, like air resistance, are acting on the object. In the case of the water balloon, because we've ignored air resistance, its motion can be considered as free fall.

When an object is in free fall, it accelerates downward due to gravity at a constant rate. We can use kinematic equations to determine the velocity, position, and time for such motion. An important thing to remember is that, irrespective of the initial velocity, once in free fall, only gravity is influencing the speed.
Gravitational Acceleration
Gravitational acceleration ( g ) is the rate at which an object speeds up as it falls freely to the ground. On Earth, this acceleration is approximately 9.81 meters per second squared (m/s²). It's important to remember this value because it's a key factor in calculating many properties of motion, such as velocity and distance fallen.

In our water balloon example, gravitational acceleration causes the balloon to accelerate as it falls. Even though it started with an initial downward velocity of 6.00 m/s, the constant pull of gravity will increase its speed steadily. This means that every second, its speed increases by 9.81 m/s due to gravity alone.
Kinematic Equations
Kinematic equations are a set of formulas used to describe the motion of objects when they move with uniform acceleration. These equations relate the five fundamental variables: initial velocity ( v_0 ), final velocity ( v ), acceleration ( a ), time ( t ), and displacement ( y ).

Here's a quick rundown of the most used kinematic equations:
  • Velocity: \( v = v_0 + at \)
  • Displacement: \( y = v_0t + \frac{1}{2}at^2 \)
  • Velocity squared: \( v^2 = v_0^2 + 2ay \)

In the water balloon scenario, these equations help calculate the balloon's velocity and distance fallen over time. Starting with initial conditions, you can plug values into these equations to find the desired information about the object's motion.
Uniform Acceleration
Uniform acceleration refers to a constant change in velocity per unit of time. In simpler terms, if an object experiences uniform acceleration, its speed will increase or decrease at a steady rate.

A prime example is the acceleration due to gravity, which stays unchanged at 9.81 m/s² for any object in free fall. This constant acceleration is what makes the calculation of falling objects predictable using kinematic equations.

In our exercise, the water balloon thrown downward initially means it starts with some speed. The addition of uniform gravitational acceleration causes a linear increase in its velocity over time, making it relatively straightforward to analyze its motion.

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Most popular questions from this chapter

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 \(b=240 \mathrm{m} / \mathrm{s}^{2}\) and \(c=0.120 \mathrm{m} / \mathrm{s}^{3}\) . (a) Calculate the average velocity of the car for the time interval \(t=0\) to \(t=10.0 \mathrm{s}\) . (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?

Sam heaves a \(16-\) - Ib shot straight upward, giving it a constant upward accleration from rest of 45.0 \(\mathrm{m} / \mathrm{s}^{2}\) for 64.0 \(\mathrm{cm} .\) He releases it 2.20 \(\mathrm{m}\) above the ground. You may ignore air resistance. \((\mathrm{a})\) What is the speed of the shot when Sam releases it? (b) How high above the ground does it go? (c) How much time does he have to get out of its way before it returns to the height of the top of his head, 1.83 \(\mathrm{m}\) above the ground?

A large boulder is ejected vertically upward from a volcano with an initial speed of 40.0 \(\mathrm{m} / \mathrm{s}\) . Air resistance may be ignored. (a) At what time after being ejected is the boulder moving at 20.0 \(\mathrm{m} / \mathrm{s}\) upward? (b) At what time is it moving at 20.0 \(\mathrm{m} / \mathrm{s}\) downward? (c) When is the displacement of the boulder from its initial position zero? (d) When is the velocity of the boulder zero? (e) What are the magnitude and direction of the acceleration while the boulder is (i) moving upward? (ii) Moving downward? (iii) At the highest point? (f) Sketch \(a_{y}-t, v_{y}-t,\) and \(y-t\) graphs for the motion.

An alert hiker sees a boulder fall from the top of a distant cliff and notes that it takes 1.30 s for the boulder to fall the last third of the way to the ground. You may ignore air resistance. (a) What is the height of the cliff in meters? (b) If in part (a) you get two solutions of a quadratic equation and you use one for your answer, what does the other solution represent?

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 \(\mathrm{s}\) , 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?

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