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liberstina [14]

4 years ago

15

A pole-vaulter is nearly motionless as he clears the bar, set 4.2 m above the ground. He then falls onto a thick pad. The top of

the pad is 80 cm above the ground, and it compresses by 50 cm as he comes to rest. What is the magnitude of his acceleration as he comes to rest on the pad?

1 answer:

Mandarinka [93]4 years ago

5 0

initial height of the Pole - vaulter = 4.2 m

now when he reached the top of the pad his height is 80 cm

so the total displacement of the person will be = 4.2 - 0.8 = 3.4 m

now we can use kinematics to find the speed just before he touch the pad

$v_f^2 - v_i^2 = 2 a d$

$v_f^2 - 0 = 2*9.8*3.4$

$v_f^2 = 66.64$

$v_f = 8.16 m/s$

now when he compressed on pad the distance after which he will stop is 50 cm

so now again using kinematics we can say

$v_f^2 - v_i^2 = 2 a d$

$0 - 8.16^2 = 2*a*0.50$

$a = -66.64 m/s^2$

so it is acceleration by -66.64 m/s^2

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Answer:

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(c) 110 m

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$v_1 = v_0 + a_1t_1 = 0 +1.76t_1 = 1.76t_1$

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