Answer :
The force exerted on his torso by his legs during the deceleration is 4365 N.
Explanation:
Mass of the torso m=45kg
Height of the building s=3.5 m
Decelerating distance=0.71 m
when he jumps to the ground, the only acceleration is acceleration due to gravity g
motion1 from top to ground
initial velocity u=0
we have to calculate final velocity v using the following equation of motion.
[tex]v^2-u^2=2gs\\v^2-0^2=2\times 9.8\times3.5=68.6\\v=\sqrt{68.6} \\=8.3[/tex]
use height of the building as the distance s as the jump from top to the ground is only described here.
Motion 2 on the ground
v=0
u=8.3(final velocity of motion 1)
The deceleration after striking the ground can be calculated from the equation of motion
[tex]v^2-u^2=2as\\\\a=v^2-u^2/2\times 0.71\\=0^2-8.3^2/0.71=97 m/s^2[/tex]
The decelerating distance is used in the place of s since since the motion after hitting the ground is described in this case.
The equation of force is
[tex]F=ma\\=45\times 97=4365 N[/tex]
The magnitude of the average force exerted on his torso by his legs during deceleration is 2,172.6 N.
The given parameters;
- height of the house, h = 3.5 m
- distance of deceleration, d = 0.71 m
- mass of the tensor, m = 45 kg;
The final speed of the person when he jumps to ground is calculated as;
[tex]\frac{1}{2} mv^2 = mgh\\\\v^2 = 2gh\\\\v = \sqrt{2gh} \\\\v = \sqrt{2\times 9.8 \times 3.5 } \\\\v = 8.28 \ m/s[/tex]
The deceleration of the torso over the given distance is calculated as;
[tex]v^2 = u^2 + 2as\\\\a = \frac{v^2 - u^2}{2s} \\\\a = \frac{0-8.28^2}{2\times 0.71} \\\\a = -48.28 \ m/s^2[/tex]
The magnitude of the average force exerted on his torso by his legs during deceleration is calculated as;
[tex]F = ma\\\\F = 45 \times 48.28\\\\F = 2,172.6 \ N[/tex]
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