Answer to Question #278237 in Mechanics | Relativity for Mishita

Question

1. A person is jumping from a height ho= 10 m with zero initial velocity and a cricket ball is thrown directly upward with an initial velocity =9.8 m/s at the same instant when time t=0 s as shown in Fig. 1. Consider h=11.5m and i and to be the unit vectors along the X and Y axis. Gravity is acting along the direction and the magnitude of gravitational acceleration is given by g 9.81 m/s². The red dot on the person represents his center of mass.

(a)If the person and the cricket ball consist of a system, find the position and velocity of the center of mass of the system at time t = 0.5 s

(b)If it takes 0.01 seconds for the person to come to complete rest when he hits the ground, calculate the average force acting on the person.

(c)Now consider when the person hits the ground, he bends his knees such that the center of mass of his body is lowered by 15 cm. If the maximum force that the bone of his legs can withstand is 50000 N, will he break his legs?

Answer 1

a)

coordinates of ball:

$x_1=20$ m

$y_1=v_0t-gt^2/2$

$y_1(0.5)=9.8\cdot0.5-0.25g/2=3.67$ m

coordinates of person center of mass:

$x_2=5$ m

$y_2=11.5-gt^2/2$

$y_2(0.5)=11.5-0.25g/2=10.27$ m

coordinates of center of mass of system:

$x=\frac{m_1x_1+m_2x_2}{m_1+m_2}=\frac{0.2\cdot20+60\cdot5}{60+0.2}=5.05$ m

$y=\frac{m_1y_1+m_2y_2}{m_1+m_2}=\frac{0.2\cdot3.67+60\cdot10.27}{60+0.2}=10.25$ m

velocity of the center of mass of the system:

$v=\frac{m_1y'_1+m_2y'_2}{m_1+m_2}$

$v_1=y'_1=v_0-gt$

$y'_1(0.5)=9.8-0.5g=4.90$ m/s

$v_2=y'_2=-gt$

$y'_2(0.5)=-0.5g=-4.91$ m/s

$v=\frac{0.2\cdot4.9-60\cdot4.91}{60+0.2}=-4.88$ m/s

b)

from energy conservation:

speed of person when he hits the ground:

$v=\sqrt{2gh}=\sqrt{2g\cdot10}=14$ m/s

Average acceleration when person hits the ground:

$a=v/0.01=1400$ m/s²

average force acting on the person:

$F=ma=60\cdot1400=84000$ N

c)

final speed of centre of mass:

$v=\sqrt{2g(h+0.15)}=\sqrt{2g\cdot10.15}=14.11$ m/s

$a_1=(14.11-14)/0.01=1100$ m/s²

$F_1=ma=60\cdot1100=66000$ N

then result force acting on the person

$F-F_1=84000-66000=18000$ N $<50000$ N

so, he will not break his legs