Problem 6.38 Part A How fast is he going as he lands on the trampoline, 2 m belo

Question

Problem 6.38 Part A How fast is he going as he lands on the trampoline, 2 m below?(Figure 1) A 61-kg trampoline artist jumps upward from the top of a platform with a vertical speed of 4.5 m/s Express your answer to two significant figures and include the appropriate units u= 1 Value Units Figure 1 of 1 Submit My Answers Give Up Part B If the trampoline behaves like a spring of spring constant 6.8×104 N/m , how far does he depress it? Express your answer to two significant figures and include the appropriate units. 2.0 m yValue Units Submit My Answers Give Up

Explanation / Answer

6.38 :

Given: m= 61kg ; v0y= 4.5 m/s

solution:

(a)

Assuming upward is positive direction.

Using the equation:

y= y0+v0yt+(1/2)gt2

-2=0+ (4.5)t+(1/2)(-9.81)t2

-4.9t2+4.5t+2=0

4.9t2-4.5t-2=0

Solving quadratic equation , we get:

t= 1.25 & t=-0.33

Time can not be a negative value hence

t= 1.25 s

Using the equation

v= v0y+gt

= 4.5 + (-9.81)(1.25)

= -7.76 m/s

negative sign indicates downward direction.

(b) By conservation of energy:

E= mgh+1/2 mv2 =1/2 kh2

61[9.81(2+h) + (0.5)(4.5)2]= (0.5)(6.8×104)h2

19.62 +9.81h+10.125 = 1114.75 h2

1114.75h2-9.81h -29.745=0

By solving we get h= 0.168& -0.159

Hence h = 0.168 m

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