I need help with part a, b, c, and d. Thank you for your help. **Problem 4: Superhero Physics Scenario**An 80-kg superhero sits at rest on top of a 5.0 meter wall as shown below. Using a rope, he swings toward a 70-kg villain who is initially at rest. The superhero releases the rope just before colliding with and clinging to the villain. After the collision, the superhero and villain slide together across the floor until they are brought to rest by friction. Determine:a. The velocity of the superhero *just before* he collides with the villain.b. The velocity of the superhero/villain system immediately following the collision.c. The impulse delivered by the floor to the combined masses.d. If the men slide across the floor over a distance of 4.75 m, apply the Work-Energy Theorem to determine the force of friction acting upon them.---**Diagram Explanation:**- The diagram shows a wall measuring 5 meters in height.- The superhero swings down from the wall using a rope.- The trajectory of the swing is indicated by a dashed line.- The point where the superhero releases the rope and collides with the villain is depicted.- After the collision, the superhero and villain slide together over a horizontal distance, marked as Δx = 4.75 meters, before coming to rest.This scenario involves principles of physics such as energy conservation, momentum, and friction.

Name: I need help with part a, b, c, and d. Thank you for your help. **Probl
Uploaded: 2024-03-20T06:57:37.000Z
Channel: Bartleby
Description: I need help with part a, b, c, and d. Thank you for your help. **Problem 4: Superhero Physics Scenario**An 80-kg superhero sits at rest on top of a 5.0 meter wall as shown below. Using a rope, he swings toward a 70-kg villain who is initially at rest

a) The reduction in potential energy = msuperhero×gh Hence the kinetic energy will also increase by the same amount. Hence the velocity v1 of superhero just before collision is given by: msuperhero×gh =12msuperherov12v1=2gh=2·9.8·5=9.89 m/s b) By conservation of momentum just before and after collision, the velocity v of hero/villain is given by: msuperhero×v1=mhero/villian×v⇒80×9.89=80+70×v⇒v=80·9.8980+70=5.275 m/sc) Note that for the sliding par, the final velocity is zero and intial velocity is v Impulse = change in momentum mhero/villian×0-mhero/villian×v=-mhero/villian×v=-80+70×5.275=-791.25 kg-m/sAs a part of Bartleby guidelines, only three subparts of a question can be answered at once. Please post for the remaining parts separately.