Ok, here is the question: A person falls 107 feet onto a box. The box was crushe

Question

Ok, here is the question:

A person falls 107 feet onto a box. The box was crushed18.7 inches. Answer the following:

a. What was the person's speed just before hitting the box? (I figured that one out. It is 25.28m/s)

b. What is the person's acceleration (m/s2) whilecrushing the box? Assume constant acceleration.

c. How long did it take (ms) for the person to come to a stopafter first contacting the box?

So I figured out a, but B & C I don't get. Accordingto the answer sheet, B is -673 and C is 37.6. I just don't know how those numbers were calculated.

I have three pages of notes where I have plugged numbers intodifferent Kinematic Equations for Motion of a Particle and usedother formulas, but nothing has given me the correct answer.

Any, help would be appreciated and I promise to rate the highestrating for a good explanation.

Thanks.

Explanation / Answer

(a) use    final speed2 = 2 * acceleration * distance .         v2 = 2 * 32 ft/s2 * 107 ft =  6848 ft2 /s2 .       v = 6848 =   82.75 ft/sec =   25.3 m/s . (b) Use the same equation... .         finalspeed2 = initial speed2 + 2 a d .       0 =   25.32   + 2 a * 0.475    (in meters) .     a = - 673 m/s2 . (c)    nowuse             final speed = initial speed + at .        0 =  25.3   + -673 * t .    t   = 25.3 / 673 =  0.0376 s =   37.6 ms . (c)    nowuse             final speed = initial speed + at .        0 =  25.3   + -673 * t .    t   = 25.3 / 673 =  0.0376 s =   37.6 ms

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