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A ball is tossed directly upward from an initial height of 5.0 m above the ground with an initial speed of 8.0 m/s. What are the maximum height, ymax, of this ball above the ground, and the time, Δt, from when the ball is launched to when it reaches the ground (if one may take the gravitational acceleration, g = 9.8 m/s^2 downward, and if one may neglect any effects due to air resistance)?

          A ball is tossed directly upward from an initial height of 5.0 m above the ground with an initial speed of 8.0 m/s. What are the maximum height, ymax, of this ball above the ground, and the time, Δt, from when the ball is launched to when it reaches the ground (if one may take the gravitational acceleration, g = 9.8 m/s^2 downward, and if one may neglect any effects due to air resistance)?
        
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University Physics with Modern Physics
University Physics with Modern Physics
Hugh D. Young 14th Edition
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A ball is tossed directly upward from an initial height of 5.0 m above the ground with an initial speed of 8.0 m/s. What are the maximum height, ymax, of this ball above the ground, and the time, Δt, from when the ball is launched to when it reaches the ground (if one may take the gravitational acceleration, g = 9.8 m/s^2 downward, and if one may neglect any effects due to air resistance)?
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Transcript

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00:01 A ball is being tossed from a height of five meters above the ground, and it has an initial velocity of 8 meters per second.
00:09 So we know the ball is going to go to its maximum height where velocity is zero and come back down with the initial velocity of zero here.
00:19 Acceleration in this case is negative 9 .8 meters per second square because it is that due to gravity, and i'm taking the downward direction as negative.
00:27 So for the first thing we want to know is what is the maximum height above the ground this reaches? and so we're going to use the equation.
00:37 Final velocity squared is equal to initial velocity squared plus two times acceleration times displacement.
00:44 And in this case, it's going to be changed in why because it's going up and down.
00:48 So subtract initial velocity from both sides.
00:52 That's in subtract.
00:58 And divide both sides by two in acceleration.
01:00 So our equation is change in position is equal to final velocity squared minus initial velocity squared divided by two times acceleration, which is gravity.
01:14 And so our change in y is equal to zero minus eight squared divided by two times negative 9 .8.
01:24 And when you do this, you get 3 .27 meters.
01:29 But that's not including the five.
01:31 We want to know above the ground.
01:32 So we add five to that.
01:33 So our total distance is 3 .27 plus 5...
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