A shot putter releases the shot some distanceabove the level ground with a velocity of 12.0 m/s,51.0 ° above the horizontal. The shot hits theground 2.08 s later. You can ignore air resistance.Part CWhat is the x-component of the shot's velocity at the beginning of its trajectory?ΠV ΑΣφm/sVæ0Previous Answers Request AnswerSubmitII A shot putter releases the shot some distanceabove the level ground with a velocity of 12.0 m/s,51.0 ° above the horizontal. The shot hits theground 2.08 s later. You can ignore air resistance.Part DWhat is the y-component of the shot's velocity at the beginning of its trajectory?m/sVyo =SubmitRequest Answer

When a projectile is launched with an initial velocity ‘u’, the sum of the horizontal and the…

Answered: A shot putter releases the shot some…

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 9P

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a physics student on planet exidor throws a ball, and it follows the parabolic trajectory shown in the figure. the ball's position is shown at 1s intervals until t = 3s. At t = 1s the ball's velocity is v = (2.0i + 2.0j)m/s...after determining the ball's velocity at t = 0. t = 2, and t = 3 s, find the BALL's LAUCH ANGLE
A batter hits a baseball so that it leaves the bat at speed Vo = 37.0m/s at an angle ao = 53.1°, at a location where g = 9.80 m/S2. (a) Find the position of the ball, and the magnitude and direction of its velocity, at t = 2.00 s. (b) Find the time when the ball reaches the highest point of its flight and find its height h at this point. (c) Find the Iwrizontal range R-that is, the horizontal distance from the starting point to where the ball hits the ground.
Eddie Manalak is a Filipino stuntman, he wishes to cross the 520m distance between two tall buildings by being shot from a cannon. He wishes to be launched at 60° to the horizontal so that he can spend more time in waving to the crowd. With what speed must he be launched? ANS. 76.7 m/s
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A projectile is projected with a velocity 9.8 m/s at 15° above the horizontal. What is the horizontal range of the projectile ? ( g = 9.8 m/s2)
A projectile is launched on earth at an angle θ, relative to horizontal direction. At half of its maximum height the speed of the projectile is 1.00 m/s, and at its maximum height the speed of the projectile is 0.41 m/s. What is the angle θ ?θ = °
An object is fired from the ground at 124.52ft/sec at an angle of 34.35deg. along the +x axis. Find the following: 1. horizontal component of initial velocity 2. VERTICAL component of initial velocity 3. After 3.56 seconds, how far has the object traveled in feet horizontally? 4. How high is the object after 3.56 seconds?
A batter hots a ball, giving it a velocity of 100 ft/s at 40 degree angle to the horizontal. One second after the ball is hit, a fielder standing in the direct path of the ball at a distance of 250 ft starts running after the ball at a constant speed. How fast must be run to catch the ball?The answer is 18.7ft/s
The trajectory of a bullet is given byx=v0(cosα)t,y=v0(sinα)t−12gt2, wherev0= 500m/s,g= 9.8m/s2, andα= 30degrees. When will the bullet hit the ground? How far from the gun willthe bullet hit the ground?
the leprechaun in the shamrock patch was named O'Malley and since he was disrupted from sitting in his patch he decided to go off and play a round of his favorite sport in where leprechaun kick a gold coin from a tee into a post 317m away. if he kicked the coin at an angle of 60 degree at a velocity of 60 m/s how high was the arch of the rainbow created by the trajectory? how long did it take to get to the pot? the officials in dublin decided to add a goal post to "kick O coin" 300 meters away and so many 33 meters high. would O'Malley's kick have cleared this goal post?
A ball is thrown with an initial velocity of (30 m/s) at an angle of (60°) above the horizontal, as shown in figure below. Find : (a) The maximum height of the ball? Answer : ( Ymax = 34.5m) (b) The time to rise to the top of the trajectory? Answer:(t,=2.65s) (@) The total time the ball is in the air? Answer: (t:=5.305)
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