anonymous
  • anonymous
A 0.145 kg baseball is thrown horizontally at 15 m/s from a platform, so that it starts at 5.2 m above the ground. How much has its kinetic energy increased when it reaches the ground? 16 J , 24 J , 0 J , or 7.4 J ? :/
Physics
  • Stacey Warren - Expert brainly.com
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SOLVED
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jamiebookeater
  • jamiebookeater
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Michele_Laino
  • Michele_Laino
the initial kinetic energy is: \[KE = \frac{1}{2}m{v^2} = \frac{1}{2} \times 0.145 \times {15^2} = ...Joules\]
anonymous
  • anonymous
so we get 16.3125? so our solution is 16 J?
Michele_Laino
  • Michele_Laino
that result is right! Nevertheless it is not our final answer!

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anonymous
  • anonymous
ohh okay! what is netx?
Michele_Laino
  • Michele_Laino
we have to consider that situation: |dw:1433103218745:dw| our ball, in free falling, has gained a vertical velocity
anonymous
  • anonymous
yes:)
Michele_Laino
  • Michele_Laino
now, we have to compute that vertical speed
anonymous
  • anonymous
ok! how do we do taht?
Michele_Laino
  • Michele_Laino
such vertical speed is given by the so called formula of Torricelli, which is: \[v = \sqrt {2gh} = \sqrt {2 \times 9.81 \times 5.2} = ...m/\sec \]
anonymous
  • anonymous
10.10006 ? what happens now? :O
Michele_Laino
  • Michele_Laino
ok! Now we have this situation: |dw:1433103512624:dw|
Michele_Laino
  • Michele_Laino
namely the final velocity has the subsequent magnitude: \[{v_f} = \sqrt {{{10}^2} + {{15}^2}} = ...m/\sec \]
anonymous
  • anonymous
18.0277?
Michele_Laino
  • Michele_Laino
ok! So the final kinetic energy is: \[KE = \frac{1}{2}m{v^2} = \frac{1}{2} \times 0.145 \times {18^2} = ...Joules\]
anonymous
  • anonymous
okay! we get 23.49 so about 24 J is our solution?
Michele_Laino
  • Michele_Laino
no, since that the requested energy change is: \[\Delta KE = 23.49 - 16 = ...Joules\]
anonymous
  • anonymous
ohh okay so we get 7.49 so about 7.4 J is our answer? :O
Michele_Laino
  • Michele_Laino
yes! that's right!
anonymous
  • anonymous
yay!! thanks@!!
Michele_Laino
  • Michele_Laino
:)

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