no rounding off A cannon ball is fired with an initial speed of 123 m/s at angle of 60 degrees from the horizontal.Express the initial velocity as a linear combination of its unit vector components.Vo = (m/s) ↑ +(m/s)At the maximum height, the speed of the cannon ball is v =m/s and the magnitude of its acceleration is a =m/s2.The time needed to reach maximum height is t =S.The maximum height reached by the cannon ball is H =m.

Hello. Since your question has multiple sub-parts, we will solve the first three sub-parts for you.…

A cannon ball is fired with an initial speed of 123 m/s at angle of 60 degrees from the horizontal. Express the initial velocity as a linear combination of its unit vector components. Vo = ( m/s) i + ( m/s) î At the maximum height, the speed of the cannon ball is v = m/s and the magnitude of its acceleration is a = m/s2. The time needed to reach maximum height is t = S.

A cannon ball is fired with an initial speed of 123 m/s at angle of 60 degrees from the horizontal. Express the initial velocity as a linear combination of its unit vector components. Vo = ( m/s) i + ( m/s) î At the maximum height, the speed of the cannon ball is v = m/s and the magnitude of its acceleration is a = m/s2. The time needed to reach maximum height is t = S.

Principles of Physics: A Calculus-Based Text

5th Edition

ISBN:9781133104261

Author:Raymond A. Serway, John W. Jewett

Publisher:Raymond A. Serway, John W. Jewett

Chapter3: Motion In Two Dimensions

Section: Chapter Questions

Problem 1P: A motorist drives south at 20.0 m/s for 3.00 min, then turns west and travels at 25.0 m/s for 2.00...

See similar textbooks

Similar questions

A projectile is launched on the Earth with a certain initial velocity and moves without air resistance. Another projectile is launched with the same initial velocity on the Moon, where the acceleration due to gravity is one-sixth as large. How does the range of the projectile on the Moon compare with that of the projectile on the Earth? (a) It is one-sixth as large. (b) It is the same. (c) It is 6 times larger. (d) It is 6 times larger. (e) It is 36 times larger.
A cannon ball is fired with an initial speed of 123 m/s at angle of 60 degrees from the horizontal. Express the initial velocity as a linear combination of its unit vector components. ( m/s) + ( m/s) At the maximum height, the speed of the cannon ball is v = ? m/s and the magnitude of its acceleration is a = ? m/s2. The time needed to reach maximum height is t = ? s. The maximum height reached by the cannon ball is H = ? m.
A football is placed at rest on the field with initial coordinates x0 = 0 and y0 = 0. The ball is then kicked and thereby given an initial velocity of magnitude v0 = 23.6 m/s and direction θ0 = 36.3o above the horizontal. Ignore the effects of air resistance. Initial horizontal component of the velocity v0x = Initial vertical component of the velocity v0y = Horizontal component of the acceleration ax = Vertical component of the acceleration ay = Enter the equation for the horizontal position versus time, x(t) = Enter the equation for the vertical position versus time, y(t) = Enter the equation for the horizontal velocity versus time, vx(t) = Enter the equation for the vertical velocity versus time, vy(t) = Enter the equation for the total velocity versus time, v(t) = Enter the equation for the angle of the velocity vector versus time, θ(t)= Enter the maximum height of the football Enter the range of the football Enter the time it takes for the football to reach its maximum…
On a spacecraft two engines fire for a time of 389 s. One gives the craft an acceleration in the x direction of ax = 3.41 m/s^2, while the other produces an acceleration in the y direction of ay = 7.34 m/s^2. At the end of the firing period, the craft has velocity components of vx = 1860 m/s and vy = 4290 m/s. Find the (a) magnitude and (b) direction of the initial velocity. Express the direction as an angle with respect to the +x axis.
The velocity of a particle moving in the x-y plane is given by (6.98i + 6.68j) m/s at time t = 7.60 s. Its average acceleration during the next 0.014 s is (4.1i + 5.4j) m/s2. Determine the velocity v of the particle at t = 7.614 s and the angle θ between the average-acceleration vector and the velocity vector at t = 7.614 s.Answers:v = ( _____ i + _____ j) m/sθ = _____ °
At t=0, a ball is thrown with an initial velocity of magnitude 15ms-1 at 75° to the horizontal (upwards). The magnitudes of the vertical and horizontal components of its initial velocity are 14.49 m/s and 3.88 m/s. The magnitude of the vertical component of velocity of the ball at t=0.5s is 9.585 m/s. The magnitude of the vertical component of velocity of the ball at t=2.0s is 5.13 m/s. Which way is the ball moving at this time? What is the magnitude of the horizontal component of velocity at t=2.0s?
A fish swimming in a horizontal plane has velocity i = (4.00 î + 1.00 ĵ) m/s at a point in the ocean where the position relative to a certain rock is i = (16.0 î − 3.20 ĵ) m. After the fish swims with constant acceleration for 15.0 s, its velocity is = (21.0 î − 6.00 ĵ) m/s. (a) What are the components of the acceleration of the fish? ax = m/s2 ay = m/s2 (b) What is the direction of its acceleration with respect to unit vector î? ° counterclockwise from the +x-axis(c) If the fish maintains constant acceleration, where is it at t = 30.0 s? x = m y = m In what direction is it moving? ° counterclockwise from the +x-axis
A fish swimming in a horizontal plane has velocity i = (4.00 î + 1.00 ĵ) m/s at a point in the ocean where the position relative to a certain rock is i = (16.0 î − 3.20 ĵ) m. After the fish swims with constant acceleration for 15.0 s, its velocity is = (21.0 î − 6.00 ĵ) m/s. (a) What are the components of the acceleration of the fish? ax = m/s2 ay = m/s2 (b) What is the direction of its acceleration with respect to unit vector î? ° counterclockwise from the +x-axis(c) If the fish maintains constant acceleration, where is it at t = 30.0 s? x = 542.5 Remember that you can treat the motions in the x and y directions separately. Each is then treated exactly as you would the one-dimensional case. m y = m In what direction is it moving? ° counterclockwise from the +x-axi
In the two-dimensional motion, what is the magnitude of the y component of the velocity of the projectile when it is at its apex of trajectory? Select one: a. Velocity is equal to the initial velocity b. Velocity is 9.81 m/s c. Velocity is equal to the x component d. Undetermined e. Velocity is 0 m/s
A skater is gliding along the ice at 2.2 m/s, when she undergoes an acceleration of magnitude 1.2 m/s2 for 3.0 s. At the end of that time she is moving at 5.8 m/s. (a) What must be the angle between the acceleration vector and the initial velocity vector?
If the vector components of the position of a particle moving in the xy plane as a function of time are x = (2.8 m/s2)t2 and y = (6.4 m/s3)t3, at what time t is the angle between the particle's velocity and the x axis equal to 45°?
A particle which moves with curvilinear motion has coordinates in millimeters which vary with the time t in seconds according to x = 4.8t2 - 6.6t and y = 1.9t2 - t3/4.0. Determine the magnitudes of the velocity v and acceleration a and the angles which these vectors make with the x-axis when t = 1.3 s.Answers: When t = 1.3 s,