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During a baseball game, a batter hits a high pop-up. if the ball remains in the air for a total of 6.0 s, how high does it rise? ( Assume Accel. = 10 m/s^2 )
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During a baseball game, a batter hits a high pop-up. if the ball remains in the air for a total of 6.0 s, how high does it rise? ( Assume Accel. = 10 m/s^2 )

User Phillis Peters
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1 Answer

4 votes
  • Time=6s
  • Vertex height's time= 6/2=3s
  • Acceleration due to gravity=10m/s^2

Apply second equation of kinematics


\\ \rm\hookrightarrow s=ut+(1)/(2)gt^2


\\ \rm\hookrightarrow s=1/2(10)(3)^2


\\ \rm\hookrightarrow s=5(9)


\\ \rm\hookrightarrow s=45m

It rises 45m

User Randomsock
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