A) a 50 kg gymnast jumps from the balance beam with a speed of 1.7m/s. she lands upright on the ground 124cm below. At what speed will she hit the ground B) On landing she bends…

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asked Mar 8, 2019 221k views

1 Answer

Jatniel · Answer 1 · 2019-03-14T12:21:05+0000

PART a)

By energy conservation we can say

PE_i + KE_i = PE_f + KE_f

mgh_1 + (1)/(2)mv_i^2 = mgh_2 + (1)/(2)mv_f^2

divide whole equation by mass "m" and plug in all given data

$9.8(1.24) + \frac{1]{2}(1.7)^2  = 9.8(0) + (1)/(2)(v)^2$

v_f = 5.21 m/s

so gymnast will reach the floor with speed 5.21 m/s

PART b)

Now the gymnast bend her knees by 0.1 m and comes to rest

so here we will have

v_f^2 - v_i^2 = 2 a d

0 - 5.21^2 = 2(a)(0.1)

a = 135.97 m/s^2

now the force on the gymnast will be

F = ma

F = 50 * 135.97 = 6798.5 N

so during landing the force will be 6798.5 N