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A blue ball is thrown upward with an initial speed of 23.2 m/s, from a height of

ID: 1919796 • Letter: A

Question

A blue ball is thrown upward with an initial speed of 23.2 m/s, from a height of 0.6 meters above the ground. 2.8 seconds after the blue ball is thrown, a red ball is thrown down with an initial speed of 7.5 m/s from a height of 29.1 meters above the ground. The force of gravity due to the earth results in the balls each having a constant downward acceleration of 9.81 m/s2. 1) What is the speed of the blue ball when it reaches its maximum height? m/s 2) How long does it take the blue ball to reach its maximum height? s 3) What is the maximum height the blue ball reaches? m 4) What is the height of the red ball 3.64 seconds after the blue ball is thrown? m 5) How long after the blue ball is thrown are the two balls in the air at the same height? 6) Which statement is true about the blue ball after it has reached its maximum height and is falling back down? The acceleration is positive and it is speeding up The acceleration is negative and it is speeding up The acceleration is positive and it is slowing down The acceleration is negative and it is slowing dow

Explanation / Answer

http://www.askmehelpdesk.com/physics/thrown-ball-547096.html

You're clearly on the right track, knowing that two variables MUST be equal to each other to find the answer.

In both cases you can compute the ball's position with the equation



I suspect you're on the right track, setting the two equations for equal to each other (so that you can solve for ).



However, there's one part of this that's a little tricky: The red ball begins it's flight 2.6 seconds after the blue ball. Since both equations have to be referenced to the same time, , you need to replace all of the 's in the equation for the red ball with (). Then when you set the positions of the two balls to be equal to each other, you should be able to solve for t, then plug back into either of the equations to solve for the height, .

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