# Physics

A ball is thrown horizontally from the top of a building.

The ball is thrown with a horizontal speed of 8.2 ms^-1. The side of the building is vertical. At point P on the path of the ball, the ball is distance x from the building and is moving at an angle of 60 degree to the horizontal. Air resistance is negligible.

a. For the ball at point P.
i. show that the vertical component of it's velocity is 14.2 ms^-1.
ii.determine the vertical distance through which ball has fallen
iii.determine the horizontal distance x.

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1. do the vertical, z, and horizontal , x , problems separately.
They are connected by location and time.
At start Vx = 8.2 and Vz = 0
There are no horizontal forces so Vx = 8.2 forever, or until a crash.
That means x = 8.2 t forever.

Now at V = 60 degrees down from horizontal Vx = |V| cos 60 still = 8.2
so
8.2 = |V| * 0.5
so then |V| = 16.4
and Vz = -|V| sin 60 = -16.4 * .866 = -14.2 sure enough (note I call z positive up so it is negative)
now the vertical problem:
at top z = 0 and Vz = 0
Vz = -g t = -9.81 t
so
-14.2 = -9.81 t
and
t = 1.45 seconds
and
z = 0 + 0 t - 4.9 t^2
z = -4.9 * (1.45)^2 = - 10.3 meters
so it fell 10.3 meters
x = 8.2 forever remember :)
x = 8.2 * 1.45 meters

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Damon
2. For Vertical Velocity:
Tanθ =x/8.2 m/s^2
X=8.2 * √ 3 = 14.2 m/s^2. It is downward, so -14.2 m/s^2

To measure the time, divide velocity by acceleration
a=v/t
t=v/a
t=-14.2 m/s / -9.8 m/s^2
t= 1.45 s

For distance, multiply velocity by time
d= v*t
d= 14.2 m/s * 1.45 s
d=20.59 m

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