Question

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 her knees through a height of 0.1m determine the average force exerted by her legs during the landing

Answer 1

PART a)

By energy conservation we can say

$PE_i + KE_i = PE_f + KE_f$

$mgh_1 + \frac{1}{2}mv_i^2 = mgh_2 + \frac{1}{2}mv_f^2$

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

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

$v_f = 5.21 m/s$

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

$F = ma$

$F = 50 \times 135.97 = 6798.5 N$

so during landing the force will be 6798.5 N