In a common but dangerous prank, a chair is pulled away as a person is moving downward to sit on it, causing the victim to land hard on the floor. suppose the victim falls by 0.50 m, the mass that moves downward is 79.0 kg, and the collision on the floor lasts 0.0750 s. what are the magnitudes of the (a) impulse and (b) average force acting on the victim from the floor during the collision?

A)247.4kgm/s b) 3299.0N

Explanation:

The potential energy of the victim = mgh where m is mass, g is acceleration due to gravity and h is height from where the victim fell

Pe = 79*9.81*0.5 = 1/2mv2 (kinetic energy of the motion) v is speed in m/s

v = √9.81 = 3.132m/s

Change in momentum of the victim = m(vi-vf)

Vi is zero and vf is the velocity of impact

Change in momentum = 79*(3.132-0) = 247kgm/s which also the impulse

B) average force x time = impulse = 247

Average force = 247/0.075 = 3299.04N

you go to your settings preference and in the end youll see sth that says "i want to delete my acc" click on it and u'll acc will be deleted but u have to wait.

in order to follow the instructions and complete the statements, i need to use data from table c of your student guide. can you think of a way to let me see your student guide for a while ? i can't answer the question without the data in table c.