A 75 kg trampoline artist jumps vertically upward from the top of a platform wit

Question

A 75 kg trampoline artist jumps vertically upward from the top of a platform with a speed of 3.7 m/s . (Figure 1)

Part A

How fast is he going as he lands on the trampoline, 2.0 m below?

Express your answer to two significant figures and include the appropriate units.

SubmitMy AnswersGive Up

Part B

If the trampoline behaves like a spring of spring constant 5.7×104 N/m , what is the distance he depress it?

Express your answer to two significant figures and include the appropriate units.

SubmitMy AnswersGive Up

Provide FeedbackContinue

Figure 1 of 1

Explanation / Answer

here m = 75 Kg u = 3.7 m/s

firstly he jump vertically upword at maximum height

where v = 0 m/s

so as we know that

v2 = u2 + 2as

s = 0 - (3.7)2 / (2 x - 9.8 )

s = 0.698 m

now from highest point to trampoline

u = 0 m/s H = 2 + 0.698 = 2.698 m

so V2 = u2 + 2aH

V2 = 0 + 2 x 9.8 x 2.698

V = 7.271 m/s Ans

(b) here K = 5.7 x 104 N/m

as we know that

V = K ( A2 - x2 ) / m

at equilibrium position x = 0

so V = KA2 / m

A2 = ( 7.271 x 75 ) / ( 5.7 x 104 )

A = 0.0978 m Ans

Related Questions

Navigate

• Browse (All)
• Browse A
• Subjects

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.