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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?

Short Answer

Expert verified
(a) 7.59 m/s; (b) 5.14 m; (c) 1.94 s.

Step by step solution

01

Analyze the given information

We have a shot put with a mass of 16 lb, which is acted upon by a constant upward acceleration of 45.0 \(\mathrm{m/s}^2\) over a displacement of 64.0 cm. The release point is at a height of 2.20 m above the ground.
02

Convert units where necessary

The displacement during acceleration is given in centimeters, so we need to convert this to meters for consistency:\[64.0 \text{ cm} = 0.64 \text{ m}\]
03

Calculate initial speed at release (Part a)

Using the kinematic equation:\[v^2 = u^2 + 2a s\]where \(u\) is the initial velocity (0 m/s), \(a = 45.0 \mathrm{m/s}^2\), and \(s = 0.64 \mathrm{m}\), we find the velocity \(v\):\[v^2 = 0 + 2(45.0)(0.64)\]\[v = \sqrt{57.6} \approx 7.59 \text{ m/s}\]The speed of the shot when released is approximately 7.59 m/s.
04

Calculate maximum height above the release point (Part b)

Using the kinematic equation for motion against gravity:\[v^2 = u^2 - 2gh\]where \(v = 0\) at the maximum height, \(u = 7.59 \text{ m/s}\), and \(g = 9.8 \mathrm{m/s}^2\), we solve for \(h\):\[0 = (7.59)^2 - 2(9.8)h\]\[h = \frac{57.648}{19.6} = 2.94 \text{ m}\]The maximum height above the release point is 2.94 m, and from the ground it is:\[2.20 + 2.94 = 5.14 \text{ m}\]
05

Calculate time to return to head height (Part c)

First, calculate the time to reach the maximum height using \(v = u + at\):\[0 = 7.59 - 9.8 t\]\[t_1 \approx 0.774 \text{ s}\]Next, calculate the total fall time from the maximum height of 5.14 m back down to 1.83 m using the equation:\[s = ut + \frac{1}{2}gt^2\]where \(s = 3.31 \) m (5.14 m - 1.83 m), and solve for total time \(t\):\[-3.31 = 0.5 \times g \times t^2\]Using the quadratic equation gives:\[t_2 \approx 1.17 \text{ s}\] The total time is \(t_1 + t_2 = 0.774 + 1.17 = 1.944 \text{ s}\). So, Sam has approximately 1.94 seconds to move out of the way.

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Key Concepts

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

Kinematic Equations
The kinematic equations are essential tools in solving problems related to motion. These equations relate different aspects of motion like displacement, velocity, acceleration, and time. In the context of projectile motion, kinematic equations help calculate quantities such as speed at release and maximum height reached by the object.
For instance, in this exercise, to calculate the initial speed at the point of release, we use the equation:
  • \[v^2 = u^2 + 2as\]
Here, \(v\) is the final velocity, \(u\) is the initial velocity (which is zero if starting from rest), \(a\) is the acceleration, and \(s\) is the displacement. Applying this equation enables us to find the velocity of a shot when released by determining how fast it moves given the provided acceleration and displacement. This example illustrates how kinematic equations are foundational to understanding motion in physics. They allow us to predict future states of moving objects based on current or past known parameters.
Acceleration
Acceleration refers to the rate at which an object's velocity changes over time. It is a vector quantity, which means it has both magnitude and direction. In this exercise, Sam applies a constant upward acceleration of 45.0 m/s² to the shot put.
When dealing with constant acceleration, certain kinematic equations apply. For example, if an object starts from rest, its increase in velocity can be calculated using the equation:
  • \[v = u + at\]
where \(v\) is the final velocity, \(u\) is the initial velocity, \(a\) is the acceleration, and \(t\) is the time.
In practical terms, acceleration determines how quickly an object can reach a specific velocity from a standstill, as seen when Sam heaves the shot put upward. The substantial upward acceleration he introduces allows the shot put to reach a respectable velocity after a short motion, which is crucial for evaluating how high and fast it will travel.
Velocity
Velocity is the speed of an object in a specific direction. It is another vector quantity, similar to acceleration. In the context of this exercise, velocity is used to determine the speed of the shot put as it leaves Sam's hand as well as its speed at other points in its motion.
We see that initially, the shot put's velocity is zero because it starts from rest. To find the velocity at the point of release, we apply the kinematic equation:
  • \[v^2 = u^2 + 2as\]
From there we derive the final velocity after traveling through a certain height. Understanding velocity helps us comprehend not only how fast something is moving but also in which direction. This comprehension is vital in predicting the future positions of the shot put as it travels upward, reaches its peak, and then comes down.
Projectile Motion
Projectile motion is the study of real-life objects moving in two dimensions under the influence of gravity. This motion is usually broken into horizontal and vertical components. However, in the given exercise, we primarily consider vertical motion due to the straight upward movement of the shot.
Key parameters in projectile motion include:
  • Initial velocity: the speed at which the object is projected.
  • Maximum height: the highest vertical point reached by the object.
  • Total time in motion: how long it takes for the object to return to the starting height.
In this scenario, the initial vertical velocity determines how high the shot put will go. As it rises, the upward velocity decreases under the influence of gravity until it reaches zero at the peak of its trajectory. This allows us to calculate the maximum height it attains. Understanding projectile motion is essential for predicting the path of objects thrown or propelled into the air, making it a crucial concept in physics.

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Most popular questions from this chapter

Entering the Freeway. A car sits in an entrance ramp to a freeway, waiting for a break in the traffic. The driver accelerates with constant acceleration along the ramp and onto the freeway. The car starts from rest, moves in a straight line, and has a speed of 20 \(\mathrm{m} / \mathrm{s}(45 \mathrm{mi} / \mathrm{h})\) when it reaches the end of the 120 \(\mathrm{m}\) -long ramp. \((\mathrm{a})\) What is the acceleration of the car? (b) How much time does it take the car to travel the length of the ramp? (c) The traffic on the freeway is moving at a constant speed of 20 \(\mathrm{m} / \mathrm{s}\) . What distance does the traffic travel while the car is moving the length of the ramp?

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Launch Failure. A \(7500-\mathrm{kg}\) rocket blasts off vertically from the launch pad with a constant upward acceleration of 2.25 \(\mathrm{m} / \mathrm{s}^{2}\) and feels no approciable air resistance. When it has reached a height of \(525 \mathrm{m},\) its engines suddenly fail so that the only force acting on it is now gravity. (a) What is the maximum height this rocket will reach above the launch pad? (b) How much time after engine failure will elapse before the rocket comes crashing down to the launch pad, and how fast will it be moving just before it crashes? (c) Sketch \(a_{y}-t, v_{y}-t,\) and \(y-t\) graphs of the rocket's motion from the instant of blast-off to the instant just before it strikes the launch pad.

A ball starts from rest and rolls down a hill with uniform acceleration, traveling 150 \(\mathrm{m}\) during the second 5.0 \(\mathrm{s}\) of its motion. How far did it roll during the first 5.0 \(\mathrm{s}\) of motion?

A turtle crawls along a straight line, which we will call the \(x\) -axis with the positive dircction to the right. The equation for the turtle's position as a function of time is \(x(t)=50.0 \mathrm{cm}+\) \((2.00 \mathrm{cm} / \mathrm{s}) t-\left(0.0625 \mathrm{cm} / \mathrm{s}^{2}\right) t^{2},(\mathrm{a})\) Find the turte's initial velocity, initial position, and initial acceleration. (b) At what time \(t\) is the velocity of the turtle zero? (c) How long after starting does it take the turtle to return to its starting point? (d) At what times \(t\) is the turtle a distance of 10.0 \(\mathrm{cm}\) from its starting point? What is the velocity (magnitude and direction) of the turtle at each of these times? (e) Sketch graphs of \(x\) versus \(t, v_{x}\) versus \(t,\) and \(a_{x}\) versus \(t\) for the time interval \(t=0\) to \(t=40 \mathrm{s}\)

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