k) What is the speed of the parcel right before it slams into the ground, in units of km/hour? (noteyou only need to convert the magnitude of your answer from part (i) to units of km/hr.) DroneParcelDroneParceltrajector of Parcelaf Rer relesetime=0: Take off!time= 8.00s: Perulrelesed!from dronaA parcel is connected to a flying drone as illustrated above. Note that the coordinate system is givenwith +y pointed in the upwards direction. Holding the parcel, the drone takes off moving straight upwith a velocity of 8.40. m/s. Note that, as usual, y = 0 at the ground.After 8.00 s, the drone accidentally releases the parcel! The parcel is in "freefall" immediately after it isreleased, as defined in Lecture 7. Assume that the magnitude of the acceleration, g = 9.81 m/s?, and thatthe drone continues with the same velocity after releasing the parcel. Ignore the effects of air-resistance.

Initial velocity (u) = 8.40 m time (t) = 8 seconds

Answered: Drone Parcel Drone Aval time=0:Take…

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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QUESTION 6 Problem For a projectile lunched with an initial velocity of vo at an angle of e (between 0 and 90°), a) derive the general expression for maximum height hmax and the horizontal range R. b) For what value of 0 gives the highest maximum height? Solution The components of vo are expressed as follows: Vinitial-x = vocos(e) Vinitial-y = vosin(e) а) 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: 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)ayt? 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 substituting, the vinitial-y expression above,…
H.W: A ball is thrown with an initial speed vgat an angle 0 below the horizontal from rooftop 44m above the ground. It lands 2sec later at a point Vox 32m from the base of the building. Find vo and 0 Ans. 0 = –37.32vo = 20. 1m/s 44 oy 32
MOTION IN TWO DIMENSION = 30.0 m/s 20.0° (**)m A projectile is thrown from (##) = 8.1 meters above the ground. With initial velocity of 30 m/s and 20 degrees above the horizontal. Find the horizontal displacement, x. Select one: O a. 60.9562 m Ob. 91.4343 m Oc. 76.1953 m Od. 38.0976 m
Given: X - coordinate: x = -5.00sinwt, where w is constant, xis meters, andt is seconds y - coordinate: y = 4.00 - 5.00coswt, where w is constant, y in meters, andt seconds What is the expression for velocity vector? O A. 5w(coswti+sinwi) O B. 5w(-coswtî+sinwî) O C. 5w(coswti-sinwj) O D. 5w(-coswti-sinwî)
The exit velocity of a baseball (its velocity as it leaves the bat) is 128 feet per second, in the direction of 30 above horizontal and directly towards the monster 30 fcot tall center field wall. If the ball was hit 4 feet above ground level, find a) Parametric equations that model the trajectory of the baseball, b) The maximum height of the baseball's fiight, c) If the center field fence is 400 feet from home plate is this a home run? d) What is the impact velocity of the ball? =) Sketch a graph representing the trajectory of the ball.
H.W: A ball is thrown with an initial speed voat an angle 0 below the horizontal from rooftop 44m above the ground. It lands 2sec later at a point Vox 32m from the base of the building. Find vo and 0 Ans. 0 = -37.32vo = 20. 1m/s 44 32
find the initial velocity and max displacement the ball can travel with different angles, the angles are 30,40and50.(change the angle to 40 then find new t and R’)and change the angle to 50 as well. the height of ball is 144.06 when it's 30 degree. the height of ball is 166.11 when it's 40 degree. the height of ball is 184.50when it's 50 degree. Length is 200mm not (170.6mm) --->(17.06cm)
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Problem For a projectile lunched with an initial velocity of vo at an angle of 0 (between 0 and 90°), a) derive the general expression for maximum height hmax and the horizontal range R. b) For what value of e gives the highest maximum height? Solution The components of vo are expressed as follows: Vinitial-x = Vocos(0) Vinitial-y = Vosin(e) 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 %3D 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)ayt 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)ay? substituting, the vinitial-y expression above, results to…

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