Compare the slant range R¡ and flight time t; for the depicted projectile with the range R and flight time t for a projectile (launched with speed vo and inclination angle a) which flies over a horizontal surface. Evaluate your four results for vo = 159 m/s and a = 31° Answers: R₁ = i R= m. t₁ = i m, t= i S S

Compare the slant range R¡ and flight time t; for the depicted projectile with the range R and flight time t for a projectile (launched with speed vo and inclination angle a) which flies over a horizontal surface. Evaluate your four results for vo = 159 m/s and a = 31° Answers: R₁ = i R= m. t₁ = i m, t= i S S

Name: Lesson: RECTANGULAR COORDINATE Direction: Do not cut the decimal place
Uploaded: 2024-03-20T06:57:47.000Z
Channel: Bartleby
Description: Lesson: RECTANGULAR COORDINATE Direction: Do not cut the decimal places while solving and in final answer. Compare the slant range R¡ and flight time ti forthe depicted projectile with the range R andflight time t for a projectile (launched withspeed

University Physics Volume 1

18th Edition

ISBN:9781938168277

Author:William Moebs, Samuel J. Ling, Jeff Sanny

Publisher:William Moebs, Samuel J. Ling, Jeff Sanny

Chapter3: Motion Along A Straight Line

Section: Chapter Questions

Problem 41P: Assume an intercontinental ballistic goes from rest to a suborbital speed of 6.50 km/s 60.0 s (the...

See similar textbooks

Similar questions

A projectile is fired vertically from point A with an initial speed of 215 ft/sec. Relative to an observer located at B, at what times (find two times t1 < t2) will the line of sight to the projectile make an angle of 23° with the horizontal? Compute the magnitude of the speed of the projectile at each time, and ignore the effect of aerodynamic drag on the projectile. Calculate the height h when the projectile is at point C.
The range of a projectile fired at an angle ? with the horizontal and with an initial velocity of v0 feet per second is r = 1 32 v02 sin(2?) where r is the horizontal distance (in feet) the projectile travels. An athlete throws a javelin at 75 feet per second. At what angle must the athlete throw the javelin so that the javelin travels 145 feet? (Assume ? < 45°. Round your answer to two decimal places.)? = °
Given a 5 meter building is away from you and it forms a right triangle with theta being 60 degrees from where you're standing and you throw a rock at 30.0 m/s how much time does it take to hit the building? how far will the rock fall in the time found? (treat as a free fall) at what height will the rock hit the building?
A projectile is fired from a cliff 300mhigh with an initial velocity of 400 m/s. If the firing angle is 300from the horizontal, compute the horizontal range of the projectile. (Ans. 15.74km)
SOLVE STEP BY STEP IN DIGITAL FORMAT In the following problem, break down the procedure by placing: Data, Formula / Clearances, Operations and Result If Captain Marvel throws Spiderman against Thanos to remove his gauntlet, with a throw of 800 meters, with what speed should the tour start? Consider that the Captain uses a 50º angle, measured with her digital protractor. What was her flight time and what maximum height did she reach in her trajectory?
An archer fish spies a meal of a grasshopper sitting on a long stalk of grass at the edge of the pond in which he is swimming. The fish is to successfully spit at and strike the grasshopper, which is 0.200 m away horizontally and 0.435 m above his mouth. What angle (in degrees) above the horizontal must he spit? Assume vfy = 0.
A projectile is fired from an angle of 35degrees above the horizontal with a speed of 180m/s. Calculate the height to which it rises, the time of flight, and the horizontal range.
A ship bearing N 30-degrees W moves 3 hours at 17 miles per hour. It then turns 50 degrees starboard (to the right) to continue for 4 hours at 21 miles per hour. Finally the ship turns to a bearing of S 20-degrees E and travels for 1 hour at 12 miles per hour. Where is the ship's final location relative to its start?
The cannon on a battleship ship can fire a shell a maximum distance of 39.0km What's the velocity of the shell? What maximum height does it reach? How many meters lower will its surface be 39 from the ship along a horizontal line parallel to the surface of the ship? Does your answer imply that error introduced by the assumption of a flat earth in projectile motion is significant here?
Assume an intercontinental ballistic goes from rest to a suborbital speed of 6.50 km/s 60.0 s (the actual speeds and time are classified). What is its average acceleration in meters per second and in multiples of g(9.80m/s2)
Check 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.
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?