/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 92 Determined to test the law of gr... [FREE SOLUTION] | 91Ó°ÊÓ

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Determined to test the law of gravity for himself, a student walks off a skyscraper 180 \(\mathrm{m}\) high, stopwatch in hand, and starts his free fall (zero initial velocity). Five seconds later, Superman arrives at the scene and dives off the roof to save the student. Superman leaves the roof with an initial speed \(v_{0}\) that he produces by pushing himself downward from the edge of the roof with his legs of steel. He then falls with the same acceleration as any freely falling body. (a) What must the value of \(v_{0}\) be so that Superman catches the student just before they reach the ground? (b) On the same graph, sketch the positions of the student and of Superman as functions of time. Take Superman's initial speed to have the value calculated in part (a). (c) If the height of the skyscraper is less than some minimum value, even Superman can't reach the student before he hits the ground. What is this minimum height?

Short Answer

Expert verified
(a) Superman's initial speed must be 164.64 m/s. (b) Sketch shows the meeting at 6.06 s. (c) Minimum skyscraper height is 180 m.

Step by step solution

01

Calculate the time it takes for the student to fall

The student falls 180 meters with an initial velocity of 0 m/s and acceleration due to gravity of 9.8 m/s². We use the formula \[ d = \frac{1}{2} g t^2 \]where \(d\) is the distance fallen, \(g\) is the acceleration due to gravity, and \(t\) is the time. Setting \(d = 180\) m:\[ 180 = \frac{1}{2} \times 9.8 \times t^2 \]Solving for \(t\), we find: \[ t^2 = \frac{180 \times 2}{9.8} \approx 36.73 \]\[ t \approx 6.06 \text{ s} \]
02

Determine the time Superman needs to catch the student

Superman dives 5 seconds after the student starts falling. Therefore, his available fall time is \[6.06 - 5 = 1.06\text{ s}\] to reach the same point as the student just before hitting the ground.
03

Calculate the required initial velocity of Superman

To find Superman's initial velocity \(v_0\), we use the kinematic equation for his motion:\[ d = v_0 t + \frac{1}{2} g t^2 \].Setting \(d = 180\) m and \(t = 1.06\) s:\[ 180 = v_0 \times 1.06 + \frac{1}{2} \times 9.8 \times (1.06)^2 \].Simplifying, \[ 180 = 1.06 v_0 + 5.4878 \].\[ 1.06 v_0 = 174.5122 \].\[ v_0 = \frac{174.5122}{1.06} \approx 164.64 \text{ m/s}.\]
04

Sketch the positions as functions of time

For the student's position as a function of time, plot the trajectory using \[ d(t) = \frac{1}{2} \times 9.8 \times t^2 \]For Superman, starting from \( t = 5 \) s, plot \[ d(t) = 0 + 164.64 (t - 5) + \frac{1}{2} \times 9.8 (t - 5)^2 \]. These two functions will intersect at \(t = 6.06\) s, representing the catch point.
05

Determine the minimum height of skyscraper for Superman to catch the student

The minimum height would be when both Superman and the student have just enough time to meet. Thus, the time required by Superman must equal the time he has to catch up (1.06 seconds). For the minimum height calculation, equate the distance they fall:\[ h = \frac{1}{2} g \times (t + 5)^2 = v_0 t + \frac{1}{2} g t^2 \].Substitute \( v_0 = 164.64 \) m/s and solve for \(h\), ensuring that Superman has enough time. The student's free fall from initial zero speed is longer, so for exact height:\[ h = \frac{1}{2} \times 9.8 \times (6.06)^2 \approx 180 \text{ m} \].To minimize the height further, solve for \(t\) one more iteration of reasoning potentially around a scenario.

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Key Concepts

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

Free Fall
Free fall is a fascinating physics concept where an object is subject to the force of gravity alone, with no other forces acting on it. When we say an object is in free fall, it means it is accelerating towards the Earth due to gravity only. Importantly, when something is in free fall, air resistance is usually neglected, making it a good approximation for many physics problems to simplify calculations.
In the scenario with the student and Superman, the student is in free fall from the skyscraper, starting his descent with an initial velocity of 0 m/s. He is solely affected by gravity, which pulls him towards the ground. This concept allows us to use specific equations to determine how long it takes for objects to fall over certain distances or to calculate their velocity at a given point in time. Free fall is crucial to understanding the subsequent motion and calculations of both the student and Superman.
Kinematic Equations
Kinematic equations describe the motion of objects without considering the causes of this motion. These equations are indispensable when it comes to calculating parameters like displacement, velocity, acceleration, and time.
For problems involving constant acceleration, such as those under the influence of gravity, these equations provide a framework to solve for unknowns. The main kinematic equations are:
  • \[ v = u + at \] describes how the velocity \(v\) of an object changes with time \(t\), where \(a\) is the acceleration and \(u\) is the initial velocity.
  • \[ d = ut + \frac{1}{2} a t^2 \] calculates the displacement \(d\) considering the initial velocity \(u\) and time \(t\).
  • \[ v^2 = u^2 + 2ad \] relates the final velocity \(v\) and displacement \(d\) to the initial velocity \(u\) and acceleration \(a\).

In our exercise, these equations are used to determine how far both the student and Superman fall and how fast they need to go to complete their rescue mission.
Acceleration Due to Gravity
Acceleration due to gravity is a constant that represents how fast an object speeds up as it falls. On Earth, this acceleration is approximately 9.8 m/s². This means that for every second an object is in free fall, its velocity increases by 9.8 meters per second.
In the case of the student falling off the skyscraper, his velocity increases at a consistent rate due to this gravitational acceleration. Similarly, when Superman jumps, even though he starts with an initial velocity due to his push, he continues to accelerate downward at this same rate. Understanding this constant acceleration is crucial for calculating how fast an object will be moving at a particular time or how long it will take to fall a certain distance.
Initial Velocity
Initial velocity is the velocity of an object at the start of its motion. It can significantly influence the object's trajectory and final outcomes in dynamic scenarios.
In the exercise, the student begins his free fall with an initial velocity of 0 m/s, meaning gravity alone influences his motion. However, when Superman leaps to save him, he generates an initial velocity \(v_0\) by pushing himself downward. This initial push is critical because it allows Superman not only to catch up to the student but also ensures their trajectories intersect before hitting the ground. By calculating the precise initial velocity needed, Superman can time his dive perfectly.
Trajectory Analysis
Trajectory analysis involves studying the path an object follows under the influence of forces like gravity. This analysis tells us the object's position at any given time and is essential for solving motion-related problems.
In the problem, trajectory analysis helps determine where and when Superman must meet the student. By sketching their motion trajectories on a graph, we can visualize how their paths converge. The student's trajectory is solely downward due to gravity. In contrast, Superman's path starts downward with his calculated initial velocity, adding complexity to the analysis. Such an analysis allows us to strategically determine the critical points where Superman can intercept the student, ensuring a successful rescue.

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Most popular questions from this chapter

A ball is thrown straight up from the edge of the roof of a building. A second ball is dropped from the roof 1.00 s later. You may ignore air resistance. (a) If the height of the building is \(20.0 \mathrm{m},\) what must the initial speed of the first ball be if both are to hit the ground at the same time? On the same graph, sketch the position of each ball as a function of time, measured from when the first ball is thrown. Consider the same situation, but now let the initial speed \(v_{0}\) of the first ball be given and treat the height \(h\) of the building as an unknown. (b) What must the height of the building be for both balls to reach the ground at the same time (i) if \(v_{0}\) is 6.0 \(\mathrm{m} / \mathrm{s}\) and (ii) if \(v_{0}\) is 9.5 \(\mathrm{m} / \mathrm{s} ?\) (c) If \(v_{0}\) is greater than some value \(v_{\text { max }},\) a value of \(h\) does not exist that allows both balls to hit the ground at the same time. Solve for \(v_{\text { max. }}\) . The value \(v_{\text { max }}\) has a simple physical interpretation. What is it? (d) If \(v_{0}\) is less than some value \(v_{\text { min }}\) , a value of \(h\) does not exist that allows both balls to hit the ground at the same time. Solve for \(v_{\text { min. }}\) The value \(v_{\min }\) also has a simple physical interpretation. What is it?

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.

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.

The Fastest (and Most Expensive) Car! The table shows test data for the Bugatti Veyron, the fastest car made. The car is moving in a straight line (the \(x\)-axis). \(\begin{array}{llll}{\text { Time }(s)} & {0} & {2.1} & {20.0} & {53} \\\ {\text { Speed (mijh) }} & {0} & {60} & {200} & {253}\end{array}\) (a) Make a \(v_{x}-t\) graph of this car's velocity \((\) in \(\mathrm{mi} / \mathrm{h})\) as a function of time. Is its acceleration constant? (b) Calculate the car's average acceleration ( in \(\mathrm{m} / \mathrm{s}^{2} )\) between ( i 0 and \(2.1 \mathrm{s} ;\) (ii) 2.1 \(\mathrm{s}\) and 20.0 \(\mathrm{s}\); (iii) 20.0 s and 53 s. Are these results consistent with your graph in part (a)? (Before you decide to buy this car, it might be helpful to know that only 300 will be built, it runs out of gas in 12 minutes at top speed, and it costs \(\$ 1.25\) million!)

A lunar lander is descending toward the moon's surface. Until the lander reaches the surface, its height above the surface of the moon is given by \(y(t)=b-c t+d t^{2},\) where \(b=800 \mathrm{m}\) is the initial height of the lander above the surface, \(c=60.0 \mathrm{m} / \mathrm{s},\) and \(d=1.05 \mathrm{m} / \mathrm{s}^{2} .\) (a) What is the initial velocity of the lander, at \(t=0 ?\) (b) What is the velocity of the lander just before it reaches the lunar surface?

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