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Sagot :
The velocity of the stuntman, once he has left the cannon is 5 m/s.
The right option is O A. 5 m/s
The Kinetic energy of the stuntman is equal to the elastic potential energy of the spring.
Velocity:
This is the ratio of displacement to time. The S.I unit of Velocity is m/s. The velocity of the stuntman can be calculated using the formula below.
⇒ Formula:
- mv²/2 = ke²/2
- mv² = ke².................. Equation 1
⇒ Where:
- m = mass of the stuntman
- v = velocity of the stuntman
- k = force constant of the spring
- e = compression of the spring
⇒ Make v the subject of the equation
- v = √(ke²/m)................. Equation 2
From the question,
⇒ Given:
- m = 48 kg
- k = 75 N/m
- e = 4 m
⇒ Substitute these values into equation 2
- v = √[(75×4²)/48]
- v = √25
- v = 5 m/s.
Hence, The velocity of the stuntman, once he has left the cannon is 5 m/s.
The right option is O A. 5 m/s
Learn more about velocity here: https://brainly.com/question/10962624
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