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Chapter 2: Problem 50

(I) Estimate \((a)\) how long it took King Kong to fall straight down from the top of the Empire State Building \((380 \mathrm{~m}\) high), and \((b)\) his velocity just before "landing."

Short Answer

Expert verified
King Kong falls for about 8.81 seconds and hits the ground at 86.34 m/s.

Step by step solution

01

Understanding the Problem

We need to calculate the time it takes for King Kong to fall from a height of 380 meters and find his velocity just before he hits the ground. This is essentially a free-fall problem without air resistance, using gravitational acceleration.
02

Equation for Free Fall

We'll use the equation for distance under constant acceleration: \[ s = ut + \frac{1}{2}at^2 \]where \( s = 380 \) m is the distance, \( u = 0 \) m/s is the initial velocity, \( a = 9.8 \) m/s虏 is the acceleration due to gravity, and \( t \) is the time in seconds.
03

Solve for Time (t)

Since the initial velocity \( u = 0 \), the equation simplifies to:\[ 380 = \frac{1}{2} \times 9.8 \times t^2 \]Solving for \( t^2 \):\[ t^2 = \frac{380}{4.9} \]\[ t^2 \approx 77.55 \]\[ t \approx \sqrt{77.55} \approx 8.81 \text{ seconds} \]
04

Equation for Final Velocity

We'll use the equation for finding final velocity in free fall:\[ v = u + at \]where \( u = 0 \) m/s, \( a = 9.8 \) m/s虏, and \( t \approx 8.81 \) s.
05

Solve for Velocity (v)

Plug the values into the equation:\[ v = 0 + 9.8 \times 8.81 \]\[ v \approx 86.338 \text{ m/s} \]
06

Conclusion

King Kong takes approximately 8.81 seconds to fall, and his velocity just before landing is approximately 86.34 m/s.

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Key Concepts

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

Acceleration due to Gravity
In the context of free fall, the acceleration due to gravity is a crucial factor. This constant, often represented by the symbol \( g \), has a value of approximately 9.8 meters per second squared (m/s虏) on Earth. This means that any object, in a vacuum, will increase its velocity by 9.8 m/s every second during free fall.
  • This value is derived from the gravitational force Earth exerts on objects.
  • Air resistance is typically ignored in basic calculations, allowing an ideal free-fall scenario.
  • The formula \( a = g \) signifies that gravity is the sole accelerator when no other forces act upon the object.
Understanding \( g \) helps predict how fast and how long an object will take to reach the ground. Thus, it's pivotal in calculating both time of descent and final velocity.
Final Velocity
Final velocity is the speed an object hits just before stopping or reaching an obstacle. In free fall, it can be calculated even when the descent begins from rest. For solving problems, the equation for final velocity \( v = u + at \) is used.- Here, \( u \) is the initial velocity, which in the case of starting from rest is 0 m/s.- \( a \) is the acceleration, given as 9.8 m/s虏.- And \( t \) is the time of the fall.The increase in velocity under gravity means its final velocity before impact can be quite large, such as the 86.34 m/s calculated for King Kong's fall. It gives insight into how speedily an object is moving the instant before collision, a critical factor in understanding impact forces.
Time of Descent
Time of descent in a free fall scenario refers to the duration an object takes to fall from a certain height to ground level, assuming constant gravitational pull. Using the formula for distance under constant acceleration, \( s = ut + \frac{1}{2}at^2 \), and considering that the initial velocity \( u \) is zero, we can simplify it to:\[ s = \frac{1}{2}gt^2 \]- By rearranging, you solve for \( t \) as follows:\[ t^2 = \frac{2s}{g} \]- Then take the square root to find \( t \).Time calculated, such as the 8.81 seconds for King Kong鈥檚 fall, is essential to determining the duration an object is influenced by gravity alone. This understanding can be applied to project practical scenarios like how long a ball takes to hit the ground when dropped.

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Most popular questions from this chapter

(III) A rock is thrown vertically upward with a speed of \(12.0 \mathrm{~m} / \mathrm{s} .\) Exactly \(1.00 \mathrm{~s}\) later, a ball is thrown up vertically along the same path with a speed of \(18.0 \mathrm{~m} / \mathrm{s}\). \((a)\) At what time will they strike each other? \((b)\) At what height will the collision occur? (c) Answer \((a)\) and \((b)\) assuming that the order is reversed: the ball is thrown \(1.00 \mathrm{~s}\) before the rock.

You are traveling at a constant speed \(v_{\mathrm{M}},\) and there is a car in front of you traveling with a speed \(v_{\mathrm{A}}\). You notice that \(v_{M}>v_{\mathrm{A}},\) so you start slowing down with a constant acceleration \(a\) when the distance between you and the other car is \(x\). What relationship between \(a\) and \(x\) determines whether or not you run into the car in front of you?

(III) A toy rocket moving vertically upward passes by a 2.0-m-high window whose sill is \(8.0 \mathrm{~m}\) above the ground. The rocket takes \(0.15 \mathrm{~s}\) to travel the \(2.0 \mathrm{~m}\) height of the window. What was the launch speed of the rocket, and how high will it go? Assume the propellant is burned very quickly at blastoff.

(II) An airplane travels \(3100 \mathrm{~km}\) at a speed of \(720 \mathrm{~km} / \mathrm{h}\), and then encounters a tailwind that boosts its speed to \(990 \mathrm{~km} / \mathrm{h}\) for the next \(2800 \mathrm{~km}\). What was the total time for the trip? What was the average speed of the plane for this trip? [Hint: Does Eq. 2-12d apply, or not?]

A fast-food restaurant uses a conveyor belt to send the burgers through a grilling machine. If the grilling machine is \(1.1 \mathrm{~m}\) long and the burgers require \(2.5 \mathrm{~min}\) to cook, how fast must the conveyor belt travel? If the burgers are spaced \(15 \mathrm{~cm}\) apart, what is the rate of burger production (in burgers/min)?
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