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?

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 [tex]y_o[/tex], taking a time t to complete the travel. The formulas who relate the different magnitudes are

[tex]v_x=v_o[/tex]

The horizontal velocity [tex]v_x[/tex] is the same regardless of the elapsed time

[tex]v_y=gt[/tex]

[tex]x=v_ot[/tex]

[tex]\displaystyle y=y_o-\frac{gt^2}{2}[/tex]

The plate of apple pie left the counter at a speed

[tex]v_o=0.84\ m/s[/tex]

The counter is [tex]y_o=1.38\ m[/tex] high.

a.

Knowing that

[tex]y_o=1.38\ m[/tex]

We use this formula to compute t

[tex]\displaystyle y=y_o-\frac{gt^2}{2}[/tex]

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

[tex]\displaystyle 0=y_o-\frac{gt^2}{2}[/tex]

[tex]\displaystyle y_o=\frac{gt^2}{2}[/tex]

Solving for t

[tex]\displaystyle t=\sqrt{\frac{2y_o}{g}}[/tex]

[tex]\displaystyle t=\sqrt{\frac{2(1.38)}{9.8}}[/tex]

[tex]t=0.5307\ sec[/tex]

b.

[tex]x=(0.84)(0.5307)[/tex]

[tex]x=0.4458\ m[/tex]

c.

[tex]v_x=0.84\ m/s[/tex]

[tex]v_y=(9.8)(0.5307)[/tex]

[tex]v_y=5.2\ m/s[/tex]