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Chapter 2: Problem 57

Two coconuts fall freely from rest at the same time, one from a tree twice as high as the other. (a) If the coconut from the taller tree reaches the ground with a speed \(V,\) what will be the speed (in terms of \(V\) ) of the coconut from the other tree when it reaches the ground? (b) If the coconut from the shorter tree takes time \(T\) to reach the ground, how long (in terms of \(T\) ) will it take the other coconut to reach the ground?

Short Answer

Expert verified
(a) The speed of the other coconut is \(\frac{V}{2}\); (b) the time is \(\sqrt{2}T\).

Step by step solution

01

Understanding the Problem

We have two coconuts falling from different heights. Coconut A falls from a height that is twice as high as Coconut B. We need to find the speed and time relationships between Coconut A and B.
02

Analyzing the velocity

The final velocity of an object falling from a height under gravity can be found using the formula: \(v^2 = u^2 + 2gh\), where \(v\) is the final velocity, \(u\) is the initial velocity (0 in this case), \(g\) is the acceleration due to gravity, and \(h\) is the height. For Coconut A (taller tree), \(v_A = \sqrt{2g(2h)} = \sqrt{4gh} = 2\sqrt{gh}\). For Coconut B (shorter tree), \(v_B = \sqrt{2gh}\). If \(v_A = V\), then \(v_B = \frac{V}{2}\).
03

Analyzing the time

The time an object takes to fall from a height under gravity is given by: \(t = \sqrt{\frac{2h}{g}}\). For Coconut B, this time is \(T = \sqrt{\frac{2h}{g}}\). For Coconut A, \(t_A = \sqrt{\frac{2(2h)}{g}} = \sqrt{\frac{4h}{g}}\). Therefore, \(t_A = \sqrt{2}T\).
04

Summary of the Solutions

For part (a), if the coconut from the taller tree reaches the ground with a speed \(V\), then the speed of the coconut from the shorter tree will be \(\frac{V}{2}\). For part (b), if the coconut from the shorter tree takes time \(T\) to reach the ground, then the other coconut will take time \(\sqrt{2}T\) to reach the ground.

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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 deals with the motion of objects, without considering the forces that cause these movements. In the context of free fall physics, kinematics helps us understand how objects like coconuts fall from a height and reach the ground. When observing a falling object, we are interested in factors such as velocity and time of flight.
  • This exercise involves two coconuts dropping from different heights, giving a practical example of kinematics.
  • Key kinematic equations enable us to calculate velocities and descent times by considering constant acceleration due to gravity.
Understanding the kinematic equations allows students to predict motion under simplified conditions, which can later be adapted to more complex real-world movements.
Gravitational Acceleration
Gravitational acceleration is a crucial concept when analyzing free fall. It is the acceleration at which an object is pulled toward the center of the Earth due to gravity. In kinematics calculations, we usually denote it by the symbol "g" and assume an average value of approximately 9.8 m/s虏 on Earth.
  • When an object falls freely, the only force acting on it is gravity, causing it to accelerate downwards at this constant rate.
  • In our example, gravitational acceleration makes the coconuts increase their velocity as they descend.
This implies that even if objects fall from different heights, like the coconuts, the gravitational acceleration remains constant, but the total velocity attained and time taken will differ based on the original height.
Velocity Calculation
Velocity calculation is fundamental in understanding how fast an object moves in a particular direction. In free fall scenarios, we often calculate the final velocity just before the object hits the ground, using the formula:\(v^2 = u^2 + 2gh\).
  • For our coconuts problem, the initial velocity \(u\) is 0, simplifying the equation to \(v = \sqrt{2gh}\).
  • Coconut A and Coconut B have different final velocities due to the difference in fall height.
If Coconut A's final velocity is \(V\), then Coconut B's velocity is \(\frac{V}{2}\), illustrating how height influences speed in free fall.
Time of Descent
The time of descent is the duration it takes for an object to fall to the ground. This is determined using the kinematic formula: \(t = \sqrt{\frac{2h}{g}}\).
  • This formula shows that the fall time is related square-root-wise to the height from which the object falls.
  • For Coconut B, this descent time is \(T\), and for Coconut A, the time is \(\sqrt{2}T\), due to its greater height.
Understanding this helps comprehend how the descent time and height are linked, highlighting that even a small increase in height significantly impacts the fall time.
Physics Problem Solving
Physics problem-solving often involves breaking down complex problems into simple, understandable parts. This systematic approach is evident in solving problems like the falling coconuts, where we apply fundamental physics concepts step-by-step to reach a solution.
  • Identify what is known: Initial conditions, like starting from rest.
  • Apply appropriate physics principles, such as gravitational acceleration.
  • Utilize kinematic formulas to compute velocities and times.
By honing these skills, students become better at tackling a wide range of physics problems, developing critical thinking and analytical prowess.

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Most popular questions from this chapter

Two bicyclists start a sprint from rest, each riding with a constant acceleration. Bicyclist \(A\) has twice the acceleration of bicyclist \(B ;\) however, bicyclist \(B\) rides for twice as long as bicyclist \(A\). What is the ratio of the distance traveled by bicyclist \(A\) to that traveled by bicyclist \(B\) ? What is the ratio of the speed of bicyclist \(A\) to that of bicyclist \(B\) at the end of their sprint?

A car is traveling at \(60 \mathrm{mi} / \mathrm{h}\) down a highway. (a) What magnitude of acceleration does it need to have to come to a complete stop in a distance of \(200 \mathrm{ft} ?\) (b) What acceleration does it need to stop in \(200 \mathrm{ft}\) if it is traveling at \(100 \mathrm{mi} / \mathrm{h} ?\)

The drivers of two cars having equal speeds hit their brakes at the same time, but car \(A\) has three times the acceleration of car \(B\). (a) If car \(A\) travels a distance \(D\) before stopping, how far (in terms of \(D)\) will car \(B\) go before stopping? (b) If car \(B\) stops in time \(T\), how long (in terms of \(T\) ) will it take for car \(A\) to stop?

A hot-air balloonist, rising vertically with a constant speed of \(5.00 \mathrm{~m} / \mathrm{s},\) releases a sandbag at the instant the balloon is \(40.0 \mathrm{~m}\) above the ground. (See Figure \(2.54 .)\) After it is released, the sandbag encounters no appreciable air drag. (a) Compute the position and velocity of the sandbag at \(0.250 \mathrm{~s}\) and \(1.00 \mathrm{~s}\) after its release. (b) How many seconds after its release will the bag strike the ground? (c) How fast is it moving as it strikes the ground? (d) What is the greatest height above the ground that the sandbag reaches? (e) Sketch graphs of this bag's acceleration, velocity, and vertical position as functions of time.

If the aorta (diameter \(d_{\mathrm{a}}\) ) branches into two equal-sized arteries with a combined area equal to that of the aorta, what is the diameter of one of the branches? A. \(V d_{\text {a }}\) B. \(d_{\text {a }} / \sqrt{2}\) C. \(2 d_{\mathrm{a}}\) D. \(d_{\mathrm{a}} / 2\)
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