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) î + ( 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/s?. The time needed to reach maximum height is t = S. The maximum height reached by the cannon ball is H = m.
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- 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.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.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 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?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…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 = _____ °
- In the figure, a ball is launched with a velocity of magnitude 7.00 m/s, at an angle of 41.0° to the horizontal. The launch point is at the base of a ramp of horizontal length d1 = 6.00 m and height d2 = 3.60 m. A plateau is located at the top of the ramp. Does the ball land on the ramp or the plateau? When it lands, what are the magnitude and angle of its displacement from the launch point?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-axisA 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
- 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 = (12.0 î − 2.60 ĵ) m. After the fish swims with constant acceleration for 15.0 s, its velocity is = (25.0 î − 1.00 ĵ) m/s. (a) What are the components of the acceleration of the fish? ax = ay = (b) What is the direction of its acceleration with respect to unit vector î? Draw coordinate axes on a separate piece of paper, and then add the acceleration vector with its tail at the origin. Write the numerical values for the x and y components and then use this drawing to determine the angle.° counterclockwise from the +x-axis(c) If the fish maintains constant acceleration, where is it at t = 28.0 s? x = m y = m In what direction is it moving? ° counterclockwise from the +x-axisA stone is catapulted at time t 0, with an initial velocity of magnitude 20.0 m/s and at an angle of 40.0° above the horizontal. What are the magnitudes of the (a) horizontal and (b) vertical components of its displacement from the catapult site at t =1.10 s? Repeat for the (c) horizontal and (d) vertical components at t =1.80 s, and for the (e) horizontal and (f) vertical components at t = 5.00 s.