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DaniilM [7]
2 years ago
14

When jumping straight down, you can be seriously injured if you land stiff-legged. One way to avoid injury is to bend your knees

upon landing to reduce the force of the impact. A 73.0 kg man just before contact with the ground has a speed of 6.46 m/s. In a stiff-legged landing he comes to a halt in 2.07 ms. Calculate the average net force that acts on him during this time
Physics
1 answer:
Alex Ar [27]2 years ago
7 0

Answer:

Explanation:

The man comes to halt due to reaction force acting on him in opposite direction . If R be the reaction force

impulse by net  force = change in momentum

Net force = R - mg , mg is weight of the man .

( R-mg ) x 2. 07 x 10⁻³ = 73 x 6.46 - 0

R - mg = 227.81 x 10³

Average net force = 227.81 x 10³ N .

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Where does the energy that is used to ride a bicycle up a hill come from and how is it classified?
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Frim the castle wall 20 m high shot an arrow. The initial speed of the bow is 45 m/s directed 40 degrees above horizontal. Find
True [87]

Answer:

Range of arrow = 225.09 meter

Final horizontal velocity = 34.47 m/s

Explanation:

We have equation of motion s=ut+\frac{1}{2} at^2, where u is the initial velocity, t is the time taken, a is the acceleration and s is the displacement.

Considering the vertical motion of arrow ( up direction as positive)

   We have u = 45 sin40 = 28.93 m/s, s = -20 m, a = acceleration due to gravity = -9.8m/s^2.

   -20=28.93*t-\frac{1}{2} *9.8*t^2\\ \\ 4.9t^2-28.93t-20=0

   t = 6.53 seconds or t = -0.63 seconds

   So time = 6.53 seconds.

Considering the horizontal motion of arrow

   u = 45 cos 40 = 34.47 m/s, t = 6.53 s, a = 0m/s^2

   s=34.47*6.53+\frac{1}{2} *0*6.53^2\\ \\ s=225.09m

So range of arrow = 225.09 meter

Horizontal velocity will not change , final horizontal velocity = 34.47 m/s.

7 0
3 years ago
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