EspanolA ball is thrown upward, with an initial velocity of 44 meters per second, at an angle of 68° with respect to the horizontal. The ball is thrown from a height of 3meters off the ground.The horizontal distance x from the starting point and the height y above the ground of the ball t seconds after it is thrown are given by the parametric equationsbelow.x=(vo cos e)ty=-4.9r+(vo sin )t +hHere vo is the initial velocity, 0 is the initial angle with respect to the horizontal, and h is the initial height.Use the equations to answer the following questions.(a) When does the ball reach its maximum height?Do not round any intermediate computations. Round your answer tothe nearest hundredth.| secondsUCDETS(b) What is the maximum height of the ball?Round your answer to the nearest tenth.RP-PR|meters

Name: EspanolA ball is thrown upward, with an initial velocity of 44 meters
Uploaded: 2024-03-20T06:57:47.000Z
Channel: Bartleby
Description: EspanolA ball is thrown upward, with an initial velocity of 44 meters per second, at an angle of 68° with respect to the horizontal. The ball is thrown from a height of 3meters off the ground.The horizontal distance x from the starting point and the

given: vo=44 m/sθ = 68°h = 3mx = vocosθ t y = -4.9t2 + (vosinθ)t+h

Science

Physics

A ball is thrown upward, with an initial velocity of 44 meters per second, at an angle of 68° with respect to the horizontal. The ball is thrown from a helght of 3 meters off the ground. The horizontal distance x from the starting point and the height y above the ground of the ball t seconds after it is thrown are given by the parametric equations below. x-(", cos e)t y=-4.9+(vo sin e)r +h Here vo is the initial velocity, 0 is the initial angle with respect to the horizontal, and h is the initial height. Use the equations to answer the following questions. (a) When does the ball reach its maximum height? Do not round any intermediate computations. Round your answer to the nearest hundredth. O seconds (b) What is the maximum height of the ball? Round your answer to the nearest tenth. I meters

A ball is thrown upward, with an initial velocity of 44 meters per second, at an angle of 68° with respect to the horizontal. The ball is thrown from a helght of 3 meters off the ground. The horizontal distance x from the starting point and the height y above the ground of the ball t seconds after it is thrown are given by the parametric equations below. x-(", cos e)t y=-4.9+(vo sin e)r +h Here vo is the initial velocity, 0 is the initial angle with respect to the horizontal, and h is the initial height. Use the equations to answer the following questions. (a) When does the ball reach its maximum height? Do not round any intermediate computations. Round your answer to the nearest hundredth. O seconds (b) What is the maximum height of the ball? Round your answer to the nearest tenth. I meters

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)...

See similar textbooks

Similar questions

fill in the table from the projectile drawn, be CLEAR about what in the graph goes where
A golf ball is hit off a tee at the edge of a cliff. Its x and y coordinates as functions of time are given by x = 17.0t and y = 4.20t − 4.90t2, where x and y are in meters and t is in seconds. (a) Write a vector expression for the ball's position as a function of time, using the unit vectors î and ĵ. (Give the answer in terms of t.) = m By taking derivatives, do the following. (Give the answers in terms of t.) (b) obtain the expression for the velocity vector as a function of time = m/s(c) obtain the expression for the acceleration vector as a function of time = m/s2(d) Next use unit-vector notation to write expressions for the position, the velocity, and the acceleration of the golf ball at t = 3.24 s. = m = m/s = m/s2
An object has an initial velocity of 29.0 m/s at 95.0° and an acceleration of 1.90 m/s2 at 200.0°. Assume that all angles are measured with respect to the positive x-axis. (a) Write the initial velocity vector and the acceleration vector in unit vector notation. (b) If the object maintains this acceleration for 12.0 seconds, determine the average velocity vector over the time interval. Express your answer in your unit vector notation.
The jet transport Bis flying north with a velocity vg = 470 km/h when a smaller aircraft A passes underneath the transport headed in the 68' direction shown. To passengers in B, however, A appears to be flying sideways and moving east. Determine the magnitudes of the actual velocity of A and the velocity which A appears to have relative to B. 68
A golf ball is hit upward, with an initial velocity of 47 meters per second, at an angle of 25° with respect to the horizontal. The ball is hit from a height of 8 meters above the ground. The horizontal distance x from the starting point and the height y above the ground of the ball t seconds after it is hit are given by the parametric equations below. x=(v0cosθ)t y=−4.9t^2+(v0sinθ)t+h Here v0 is the initial velocity, θ is the initial angle with respect to the horizontal, and h is the initial height. Use the equations to answer the following questions. (a)How long is the ball in the air before it first touches the ground?Do not round any intermediate computations. Round your answer to the nearest hundredth. _____ seconds (b)What is the horizontal distance the ball travels before it first touches the ground?Round your answer to the nearest whole number. _____ meters
One of the popular routes out of Toronto is towards Tokyo, Japan. An airplane model used by Air Canada for this 10345 km flight is the Boeing 787-9. It has capacity for 298 passengers. When every seat is occupied, the passengers and their luggage weigh approximately 3.41x104 kg. Approximately 6.03x104 kg of fuel has to be consumed during this one-way flight. Assuming the every seat is occupied, how much fuel in kilograms is required per passenger for the entire one-way flight? (do not include units, just the number of kilograms) Answer: Check
A rocket is fired upward with a velocity (v) of 590 feet per second. find how many seconds pass before the rocket reaches the ground (s=0) again. use the quadratic equation s=-16t^2+590t.
Vector A has magnitudee 35. When added to vector B of magnitude 20, possible magnitude of the sum is: CHoices: 47 13 5 65 zero
A person going for a walk follows the path shown in the figure, where y, is the person's resultant displacement measured from the starting point? = 254 m and 0 = 56.0°. The total trip consists of four straight-line paths. At the end of the walk, what magnitude direction ° counterclockwise from the +x axis Start 100 m End 200 m 30.00 150 m
A person walks in the following pattern: 3.0 km north, then 2.1 km west, and finally 5.2 km south. (a) How far and (b) at what angle (measured counterclockwise from east) would a bird fly in a straight line from the same starting point to the same final point?
The velocity vector of a particle is given by: V = Vje-)[sin(wt)i + cos(wt)j] where, Vo = 12.1 m/s; T = 2.5 s; w = 5.2 rad/s Calculate the magnitude of the acceleration (in m/s2) at t = 7.4 s
A small metal ball is kicked from the edge of a cliff. Its x- and y-coordinates as functions of time are given by x = 19.9t and y = 3.60t - 4.90t2, where x and y are in meters and t is in seconds. (Do not include units in your answer.) (a) Write a vector expression for the ball's position as a function of time (in m), using the unit vectors î and ĵ. (Give the answer in terms of t.) r = By taking derivatives, do the following. (b) Obtain the expression for the velocity vector v as a function of time (in m/s). (Give the answers in terms of t.) V = a = (c) Obtain the expression for the acceleration vector a as a function of time (in m/s²). m/s² m r = (d) Next, use unit-vector notation to write expressions for the position, the velocity, and the acceleration of the metal ball at t = 2.76 s. (Assume the position is in m, the velocity is in m/s and the acceleration is in m/s².) ✓ = m/s a m m/s m/s²