Problem 7.24 Part A A 55-kg person lands on firm ground after jumping from a hei
ID: 1782490 • Letter: P
Question
Problem 7.24 Part A A 55-kg person lands on firm ground after jumping from a height of 2.8 m. [Hint: The average net force on him which is related to impulse, is the vector sum of gravity and the force exerted by the ground.] (Figure 1) Calculate the impulse experienced by the persorn Express your answer to two significant figures and include the appropriate units. Enter positive value if the impulse is upward and negative value if the impulse is downward Figure 1 of 1 Submit My Answers Give Up Part B Estimate the average force exerted on the person's feet by the ground if the landing is stiff legged. Assume the body moves 1.0 cm during impact. Express your answer to two significant figures and include the appropriate units. Enter positive value if the force is upward and negative value if the force is downward mg grd F F ground 1Value Units Submit My Answers Give UpExplanation / Answer
Impulse = change in momentum
v = sqroot ( 2x9.8 x 2.8) = 7.408 m/s
a) J = 55 (0- 7.408 ) =- 407.44 kg m/s
b)force = J / time
time of collision = distance / speed= ( 1x 10^-2) / (7.408) = 0.135 x 10^-2 sec
force = 407.44/ 0.135 x 10^-2= 3018. 1 x 10 ^ 2 N
c) time= 50 x 10^-2/ 7.408 = 6.75 x 10^-2 sec
force= 407.44/ 6.75 x 10^-2= 6036.148 N apprx
a=