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A busy waitress slides a plate of apple pie along a counter to a hungry customer sitting near the

end of the counter. The customer is not paying attention, and the plate sfides off the counter
horizontally at 0.84 m/s. The counter is 1.38 m high.
a. How long does it take the plate to fall to the floor?
b. How far from the base of the counter does the plate hit the floor?
c. What are the horizontal and vertical components of the plate's velocity just before it hits
the floor?

1 Answer

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Answer:

a. 0.5307 sec

b. 0.4458 m

c. =
v_x=0.84\ m/s\ ,\ v_y=5.2\ m/s

Step-by-step explanation:

Horizontal Motion

It describes the dynamics of an object thrown horizontally in free air. The initial horizontal velocity is maintained all the time since no horizontal forces are acting. The initial vertical velocity is zero at launch time, but it grows downwards powered by the acceleration of gravity.

The object hits the ground at a distance x from the point of launching, after having traveled a vertical distance
y_o, taking a time t to complete the travel. The formulas who relate the different magnitudes are


v_x=v_o

The horizontal velocity
v_x is the same regardless of the elapsed time


v_y=gt


x=v_ot


\displaystyle y=y_o-(gt^2)/(2)

The plate of apple pie left the counter at a speed


v_o=0.84\ m/s

The counter is
y_o=1.38\ m high.

a.

Knowing that


y_o=1.38\ m

We use this formula to compute t


\displaystyle y=y_o-(gt^2)/(2)

At the moment when the plate hits the floor y=0


\displaystyle 0=y_o-(gt^2)/(2)


\displaystyle y_o=(gt^2)/(2)

Solving for t


\displaystyle t=\sqrt{(2y_o)/(g)}


\displaystyle t=\sqrt{(2(1.38))/(9.8)}


t=0.5307\ sec

b.


x=(0.84)(0.5307)


x=0.4458\ m

c.


v_x=0.84\ m/s


v_y=(9.8)(0.5307)


v_y=5.2\ m/s

User Arielle Nguyen
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