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2 years ago

8

A cup slides off a 1.1\,\text m1.1m1, point, 1, start text, m, end text high table with a speed of 1.3\,\dfrac{\text m}{\text s}

1.3 s m 1, point, 3, start fraction, start text, m, end text, divided by, start text, s, end text, end fraction to the right. We can ignore air resistance. What was the cup's horizontal displacement during the fall?

1 answer:

Tju [1.3M]2 years ago

8 0

Answer:

<h3>0.145m</h3>

Explanation:

Using the equation of motion formula y = ut + 1/2gt² where;

y is the horizontal displacement

u is the initial velocity

g is the acceleration due to gravity

t is the time

To calculate the horizontal displacement, we need to first get the time t. Using the equation:

v = u+gt

1.3² = 0+(9.8)t

1.69 = 9.8t

t = 1.69/9.8

t = 0.172s

Substituting the time into the equation above to get y

y = 0+1/2(9.8)(0.172)²

y = 4.905(0.029584)

y = 0.145m

Hence the cup's horizontal displacement during the fall is 0.145m

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