An average person can reach a maximum height of about 60 cm when jumping straight up from a crouched position. During the jump itself, the person’s body from the knees up typically rises a distance of around 60 cm. To keep the calculations simple and yet get a reasonable result, assume that the entire body rises this much during the jump. (a) With what initial speed does the person leave the ground to reach a height of 60 cm? (b) Draw a free-body diagram of the person during the jump. (c) In terms of this jumper’s weight w, what force does the ground exert on him or her during the jump?

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Chapter1: Units, Trigonometry. And Vectors
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[A1] Direction: The following questions are referring to the concept of Newton’s 2nd law and kinematics, answer it to obtain quantitative and qualitative conclusions about it. Round off your final answers in the nearest hundredths if computation is required.

  • An average person can reach a maximum height of about 60 cm when jumping straight up from a crouched position. During the jump itself, the person’s body from the knees up typically rises a distance of around 60 cm. To keep the calculations simple and yet get a reasonable result, assume that the entire body rises this much during the jump. (a) With what initial speed does the person leave the ground to reach a height of 60 cm? (b) Draw a free-body diagram of the person during the jump. (c) In terms of this jumper’s weight w, what force does the ground exert on him or her during the jump?

 

NOTES:
Newton's 2nd law states if a net external force acts on a body, the body accelerates. The
direction of the acceleration is same as the direction of the net force.
I F = ma
Equations:
E F, = T + (- w) = ma,
T = w + m = mg + ma, = mg +a,)
To determine a, rewrite the constant-acceleration equation:
%3D
vy = v0y° + 2ay(0 – y0) to:
Vy*-VOy²
a
2 (y=y0)
Transpose:
(0)*-(-10.0 m/n)*
2(-25.0 m)
+ 2. 00 m/s²
We will use T = m(g + a) in getting the Tension.
T = (800 kg)(9.80 m/s° + 2.00 m/s³) = 9440N
a =
%3!
Transcribed Image Text:NOTES: Newton's 2nd law states if a net external force acts on a body, the body accelerates. The direction of the acceleration is same as the direction of the net force. I F = ma Equations: E F, = T + (- w) = ma, T = w + m = mg + ma, = mg +a,) To determine a, rewrite the constant-acceleration equation: %3D vy = v0y° + 2ay(0 – y0) to: Vy*-VOy² a 2 (y=y0) Transpose: (0)*-(-10.0 m/n)* 2(-25.0 m) + 2. 00 m/s² We will use T = m(g + a) in getting the Tension. T = (800 kg)(9.80 m/s° + 2.00 m/s³) = 9440N a = %3!
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