14. (20 points) A shell is fired with a horizontal velocity in the positive x direction fromthe top of an 80 m high cliff. The shell strikes the ground 1330 m from the base of thecliff.a) Determine the initial speed of the shell.b) What is the speed of the shell as it hits the ground?c) What is the angle between the shell's final velocity and the horizontal?d) What is the magnitude of the displacement of the shell?

Answered: Image /qna-images/answer/d3f38cf5-b7fa-48d9-b3d9-2d55e1fa80f4.jpg

Answered: 14. (20 points) A shell is fired with a…

College Physics

11th Edition

ISBN:9781305952300

Author:Raymond A. Serway, Chris Vuille

Publisher:Raymond A. Serway, Chris Vuille

Chapter1: Units, Trigonometry. And Vectors

Section: Chapter Questions

Problem 1CQ: Estimate the order of magnitude of the length, in meters, of each of the following; (a) a mouse, (b)...

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7. A bullet is fired downward at an angle of 30° below the horizontal. 3.20 seconds later, the bullet hits a target that is 120 m horizontally to the right of the starting point. Let x direction be to the right and y direction point upward. a) Find vox, the x-component of the initial velocity of the bullet. b) Find the y-component of the initial velocity of the bullet. Voy,
1. The distance between home plate and the pitcher's mound in baseball is 60.5 ft. In MLB, baseballs are. pitched at about 90mph. Assume the ball is thrown perfectly horizontally, at 0°. How far in feet will the ball drop during its transit to home plate? Use g = 32 ft/s². (Note: this is why pitcher's mounds are raised.) 2. A pebble is tossed into the air at an angle of 62° above the horizontal, with an initial speed of 4.8 m/s. The pebble is 1.6m above the ground when it is released by the thrower's hand. a) How long will the pebble remain airborne? b) What maximum height above the ground will the pebble achieve?
1. A golf ball is hit from 4.3 m above a golfing fairway with an initial velocity of 30.0 m/s at an angle of 35° above the horizontal. a) Determine the time of flight for the ball. b) Determine the range for the golf ball. c) Determine the velocity for the golf ball the instant before the ball impacts the ground.
6) A projectile is fired at an inclination of 45° to the horizontal, with a muzzle velocity of 100 -32x? feet per second. The height h of the projectile is modeled by h(x) = + x where x 1002 is the horizontal distance of the projectile from the firing point. a. At what horizontal distance from the firing point is the height of the projectile a maximum? b. Find the maximum height of the projectile. c. At what horizontal distance from the firing point will the projectile strike the ground?
Find the magnitude of the object's velocity just before it reaches the ground.
24. A shell is fired with a horizontal velocity in the positive x direction at a speed of 330 m/s from the top of an 80- m high cliff. The shell strikes the ground 1330 m from the base of the cliff. What is the magnitude of the velocity of the shell as it hits the ground? Hint: you need to determine horizontal and vertical components of the final velocity first, from the kinematic equations, then calculate magnitude of the velocity. While the horizontal velocity remains constant, the vertical velocity does not. You will need time of flight to determine the latter. 80 m 1330 m A) 4.0 m/s B) 9.8 m/s B) 9.8 m/s C) 82 m/s D) 170 m/s (E) 332 m/s
1. A cannon is elevated at an angle of 60° with the horizontal. It fires a ball with a speed of 300 m/s from the top of a 80 m high building. a) Calculate the maximum height from the ground. b) How long is the ball in the air? c) What is the horizontal range of the ball? d) Calculate the speed at impact.
Max Height = V;*sin's (visine) 28 28 Max Ronge = v? sin 20 Vi Height 8 = 9. 81 m/sz Range 1.) A very small cannon (very small, meaning we can disregard its size) has a muzzle velocity (velocity of the projectile as it leaves the barrel) of 30 m/s and is at an angle of 35° from the ground. a.) What is the Max Height? b.) What is the Range? Somple . Diagrom Vi = inital velocty cmuzzle velocity) 0 = ongle from the grond Vi = 30 m/e Height e= 350 Ronge a) Monx Height : (30) (ain 5) Max Height = V;? sin?e 28 15- 09 m R(9.81) b.) Ronge : v.* sin 20 (30)2 sin (70*) Ronge = 86.21M * Incicode your final answers. In this case, I used a box
3. What is the value of the component in (y) of the velocity (vy), when a projectile reaches its maximum height?
Problem For a projectile lunched with an initial velocity of v0 at an angle of θ (between 0 and 90o) , a) derive the general expression for maximum height hmax and the horizontal range R. b) For what value of θ gives the highest maximum height? Solution The components of v0 are expressed as follows: vinitial-x = v0cos(θ) vinitial-y = v0sin(θ) a) Let us first find the time it takes for the projectile to reach the maximum height. Using: vfinal-y = vinitial-y + ayt since the y-axis velocity of the projectile at the maximum height is vfinal-y = _______ Then, _____ = vinitial-y + ayt Substituting the expression of vinitial-y and ay = -g, results to the following: _____= _____ -_____ t Thus, the time to reach the maximum height is tmax-height = _____/_____ We will use this time to the equation yfinal - yinitial = vinitial-yt + (1/2)ayt2 if we use the time taken to reach the maximum height, therefore, the displacement will yield the maximum height, so hmax = vinitial-yt + (1/2)ayt2…
Problem For a projectile lunched with an initial velocity of v0 at an angle of θ (between 0 and 90o) , a) derive the general expression for maximum height hmax and the horizontal range R. b) For what value of θ gives the highest maximum height? Solution The components of v0 are expressed as follows: vinitial-x = v0cos(θ) vinitial-y = v0sin(θ) a) Let us first find the time it takes for the projectile to reach the maximum height.Using: vfinal-y = vinitial-y + ayt since the y-axis velocity of the projectile at the maximum height isvfinal-y =___________ Then, ___________= vinitial-y + aytSubstituting the expression of vinitial-y and ay = -g, results to the following: __________=__________-__________t Thus, the time to reach the maximum height istmax-height = __________ /__________ We will use this time to the equation yfinal - yinitial = vinitial-yt + (1/2)ayt2 if we use the time taken to reach the maximum height, therefore, the displacement will yield the maximum height, so hmax =…
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