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masya89 [10]
2 years ago
12

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 )
Physics
1 answer:
andrew11 [14]2 years ago
5 0
  • 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+\dfrac{1}{2}gt^2

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

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

\\ \rm\hookrightarrow s=45m

It rises 45m

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Answer:

a. Final velocity, V = 2.179 m/s.  

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Explanation:

<u>Given the following data;</u>

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a. To find the velocity of the boat after it has traveled 4.75 m

Since it started from rest, initial velocity is equal to 0m/s.

Now, we would use the third equation of motion to find the final velocity.

V^{2} = U^{2} + 2aS

Where;

  • V represents the final velocity measured in meter per seconds.
  • U represents the initial velocity measured in meter per seconds.
  • a represents acceleration measured in meters per seconds square.
  • S represents the displacement measured in meters.

Substituting into the equation, we have;

V^{2} = 0^{2} + 2*0.500*4.75

V^{2} = 4.75

Taking the square root, we have;

V^{2} = \sqrt {4.75}

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b. To find the velocity if the boat has traveled 50 m.

V^{2} = 0^{2} + 2*0.500*50

V^{2} = 50

Taking the square root, we have;

V^{2} = \sqrt {50}

<em>Final velocity, V = 7.071 m/s.</em>

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