Finding Momentum when giving the mass and height

Click For Summary
SUMMARY

The discussion focuses on calculating the momentum of a 50.0-kg male dancer who leaps to a height of 0.32 m. The correct approach involves using the gravitational potential energy (PE) and kinetic energy (KE) relationship, specifically the equation mgh = ½mv², leading to the velocity formula v = √(2gh). By substituting the values for mass and height into this equation, the dancer's velocity upon reaching the ground can be determined, allowing for the calculation of momentum using P = mv.

PREREQUISITES
  • Understanding of gravitational potential energy (PE) and kinetic energy (KE) concepts
  • Familiarity with the equations of motion under constant acceleration
  • Knowledge of momentum calculation using the formula P = mv
  • Basic algebra skills for manipulating equations
NEXT STEPS
  • Study the derivation of the kinetic energy formula from gravitational potential energy
  • Learn about the conservation of energy principles in physics
  • Explore the effects of varying mass and height on momentum calculations
  • Investigate real-world applications of momentum in sports physics
USEFUL FOR

This discussion is beneficial for physics students, educators, and anyone interested in understanding the principles of momentum and energy transformations in physical activities.

Andrea Fdez
Messages
1
Reaction score
0
Moved from a technical forum thus no homework template.
A 50.0-kg male dancer leaps 0.32 m high.
(a) With what momentum does he reach the ground?

I know that to find momentum the equation is P = mv, but I only have the mass and distance. I have tried finding the time (D = 1/2at^2) and later used that time to find the velocity (v = d/t). I later used the velocity to complete my momentum equation, but my answer was incorrect. Can someone please help me understand this problem.
 
Physics news on Phys.org
v=d/t gives the velocity when the acceleration is zero. Here the acceleration is g.
 
Changing from (gravitational) PE to KE: $$mgh=½mv^2$$ from which ##v=\sqrt{2gh}##
 

Similar threads

  • · Replies 9 ·
Replies
9
Views
1K
  • · Replies 13 ·
Replies
13
Views
1K
  • · Replies 4 ·
Replies
4
Views
1K
Replies
10
Views
2K
  • · Replies 12 ·
Replies
12
Views
3K
Replies
3
Views
1K
  • · Replies 13 ·
Replies
13
Views
1K
Replies
335
Views
16K
  • · Replies 5 ·
Replies
5
Views
879
Replies
10
Views
3K