To determine the height of a flagpole, Abby throws a ball straight up and times it. She is able to determine the ball goes by the top of the pole after 0.423 s, and then reaches the top of the pole again after a total elapsed time of 1.976 s after releasing it.
What is the height of the flagpole above the launch point?
What is the height of the flagpole above the launch point?
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i recommend first knowing the 4 kinematics equation..
then use the equation vf = (vi + at) if the total time is 1.976 then the time it takes to reach it's maximum height is (1.976)/2 = 0.988. When the ball reaches its maximum height, its vf = 0. so
i will exclude the units for a while so you won't be confuse
vf = vi + at ---> 0 = vi + (-9.8)(0.988) (negative acceleration)
0 = vi - 9.6824 then vi = 9.6824 m/s
then use the equation
d = vi*t + (1/2)at^2
the substitute the value (note use t = 0.423 because it was the time elapsed to reach the pole)
d = 9.6824*0.423 + (1/2)(-9.8)(0.423^2)
d = 3.2189 meters
then use the equation vf = (vi + at) if the total time is 1.976 then the time it takes to reach it's maximum height is (1.976)/2 = 0.988. When the ball reaches its maximum height, its vf = 0. so
i will exclude the units for a while so you won't be confuse
vf = vi + at ---> 0 = vi + (-9.8)(0.988) (negative acceleration)
0 = vi - 9.6824 then vi = 9.6824 m/s
then use the equation
d = vi*t + (1/2)at^2
the substitute the value (note use t = 0.423 because it was the time elapsed to reach the pole)
d = 9.6824*0.423 + (1/2)(-9.8)(0.423^2)
d = 3.2189 meters
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but you said that 1.976 s is the total elapsed time..
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