Physics for Scientists and Engineers, Vol. 1
6th Edition
ISBN: 9781429201322
Author: Paul A. Tipler, Gene Mosca
Publisher: Macmillan Higher Education
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Chapter 6, Problem 52P
To determine
The velocity vector as function of time; to show that explicit integral of
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If we neglect air resistance, then the range of a ball (or any projectile) shot at an angle 0 with respect to the x axis and with an initial velocity vo, is given by
R(0)=Voʻ sin(20)
for 0 < e s II
g
where g is the acceleration due to gravity (9.8 meters per second per second).
For what value of 0 is the maximum range attained? (Note that the answer is numerical, not symbolic.)
Answer detailly 6
A mountain climber jumps a crevasse of width W by leapinghorizontally with speed v0. (a) If the height difference betweenthe two sides of the crevasse is h, what is the minimum value ofv0 for the climber to land safely on the other side? (b) In this case,what is the climber’s direction of motion on landing?
Chapter 6 Solutions
Physics for Scientists and Engineers, Vol. 1
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- Suppose that an artillery projectile is fired from ground level with initial velocity 3000 ft/s and initial inclination angle 40. . Assume that its air resistance deceleration is (0.0001)v2. (a) What is the range of the projectile and what is its total time of flight? What is its speed at impact with the ground? (b) What is the maximum altitude of the projectile, and when is that altitude attained? (c) You will find that the projectile is still losing speed at the apex of its trajectory. What is the minimum speed that it attains during its descent?arrow_forwardA railroad flatcar is traveling to the right at a speed of 13.0 m/srelative to an observer standing on the ground. Someone is riding amotor scooter on the flatcar. What is the velocity (magnitudeand direction) of the scooter relative to the flatcar if the scooter’svelocity relative to the observer on the ground is 18.0 m/s to theright?arrow_forward2.18. A ball is thrown vertially upward with an initial velocity v. The air resistance is proportional to the square of the velocity. Show that the velocity with which the ball returns to the original posi- tion is Vv, where v, is the terminal velocity.arrow_forward
- To decide who pays for lunch, a passengeron a moving train tosses a coin straight upward with an initialspeed of 5.25 m>s and catches it again when it returns to its initial level. From the point of view of the passenger, then, the coin’sinitial velocity is 15.25 m>s 2yn . The train’s velocity relative to theground is 112.1 m>s2xn . (a) What is the minimum speed of the coinrelative to the ground during its flight? At what point in the coin’sflight does this minimum speed occur? Explain. (b) Find the initial speed and direction of the coin as seen by an observer on theground. (c) Use the expression for ymax derived in Example 4-14 tocalculate the maximum height of the coin, as seen by an observeron the ground. (d) What is the maximum height of the coin fromthe point of view of the passenger, who sees only one-dimensionalmotion?arrow_forwardIn a later chapter, you will find that the weight of a particle varies with altitude such that w=mgr02r2where r0is the radius of Earth and ris the distance from Earth’s center. If the particle is fired vertically with velocity v0from Earth’s surface, determine its velocity as a function of position r. (Hint: use adr=vdv, the rearrangement mentioned in the text.)arrow_forwardCheck Your Understanding If the two golf shots in Example 4.9 were bunched at the same speed, which shot would have the greatest range? When we speak of the range of a projectile on level ground, we assume R is very small compared with the circumference of Earth. If, however, the range is large, Earth curves away below the projectile and the acceleration resulting from gravity changes direction along the path. The range is larger than predicted by the range equation given earlier because the projectile has father to fail than it would on level ground, as shown in Figure 4.17, which is based on a drawing in Newton’s Principia. If the initial speed is great enough, the projectile goes into orbit. Earth’s surface drops every 8000 m. In 1 s an object falls 5 m without a resistance. Thus, if an object is given a horizontal velocity of 8000 m/s (or 18,000 mi/hr) near Earth’s surface, it will go into orbit around the planet because, the surface continuously falls away from object Thu is roughly the speed of the Space Shuttle in a low Earth orbit when it was operational, or any satellite in a low Earth orbit. These and other aspects of orbital motion, such as Earth’s rotation, are covered in greater depth in Gravitation. Figure 4.17 Projectile to satellite. In each case shown here, a projectile is launched from a very high tower to avoid air resistance. With increasing initial speed, the range increases and becomes longer than t would be on level ground because Earth curves away beneath its path. With a speed of 8000 m/s, orbit is achieved.arrow_forward
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