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NARA [144]
4 years ago
5

A ball of moist clay falls 15.0 m to the ground. It is in contact with the ground for 20.0 ms before stopping. (a) What is the m

agnitude of the average acceleration of the ball during the time it is in contact with the ground?
Physics
1 answer:
rosijanka [135]4 years ago
3 0

Answer:

a= -0.86 m/s²

The negative sign shows that ball down the ground or moving down

Explanation:

Vf² - Vo² = 2gS

where

Vf = velocity of clay as it hits the ground

Vo = initial velocity of clay = 0

g = acceleration due to gravity = 9.8 m/sec^2 (constant)

S = distance travelled by clay = 15 m

Substituting appropriate values,

Vf² - 0 = 2(9.8)(15)  

Vf = 17.15 m/sec.

Formula to use is,  

V - Vf = aT

where

V = velocity of clay when it stops = 0

Vf = 17.15 m/sec (as determined above)

a = acceleration

T = 20 ms  

Put the values to find acceleration

a=(V-Vf)/T

a=(0-17.15)/20

a= -0.86 m/s²

The negative sign shows that ball down the ground

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There's a nasty wrinkle here that's kind of sneaky, and makes the work harder than it should be.

Look at the first question.  There's a number there that's dropped in so quietly that you're almost sure to miss it, but it changes the whole landscape of both of these problems.   That's where it says

" ... 20 cm mark (30 cm from the fulcrum) ... " .

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When the question talks about hanging a weight, it tells the <em>centimeter mark on the stick</em> where the weight is tied.  To solve the problem, we have to first figure out <em>how far that is from the pivot</em>, then calculate how far from the pivot to put the weight on the other side, and finally <u><em>what centimeter mark that is</em></u> on the stick.      

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= = = = = = = = = =  

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