Question

A stuntman of mass 48 kg is to be launched horizontally out of a spring-

Answer 1

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: brainly.com/question/10962624